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Asosiy kontent

Kuchsiz kislota-asos muvozanati

Kuchsiz kislota va asoslarning dissotsilanish reaksiyasi va uning muvozanat doimiysiga bogʻliqligi,   Kk va Ka. Kk va Ka ning pH ga bogʻliqligi va dissotsilanish darajasini aniqlash. 

Asosiy tushunchalar:

  • Bir negizli, kuchsiz kislotaning start text, H, K, end text tutash asosi bilan umumiy muvozanat konstantasi quyidagicha boʻladi:
K, start subscript, start text, k, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, K, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, K, end text, close bracket, end fraction
  • Kislotaning dissotsilanish konstantasi K, start subscript, start text, k, end text, end subscript kuchsiz kislotaning dissotsilanish darajasini koʻrsatadi. K, start subscript, start text, k, end text, end subscript qiymati qanchalik katta boʻlsa, kislota shunchalik kuchli va aksincha.
  • Kuchsiz asosning tutash kislotasi bilan muvozanat doimiysi ifodasi quyidagicha boʻladi:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, A, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, A, end text, close bracket, end fraction
  • Asosning dissotsilanish domiysi (yoki asosning ionlanish doimiysi) K, start subscript, start text, a, end text, end subscript kuchsiz asosning ionlanish darajasini koʻrsatadi. K, start subscript, start text, a, end text, end subscript qanchalik katta boʻlsa, asos shunchalik kuchli boʻladi va aksincha.

Kuchli va kuchsiz asos-kislotalar

Kuchli kislota va asoslar eritmada toʻliq ionlarga ajraladigan moddalar hisoblanadi. Kuchsiz kislota va asoslar esa qisman ionlanadi va bu reaksiya qaytar jarayon hisoblanadi. Kuchsiz asos va kislotalarning eritmalarida turli zaryadlangan ionlar va zaryadsiz zarrachalar dinamik muvozanat holatida boʻladi.
Ushbu mavzuda kislota va asoslarning dissotsilanish reaksiyalarini hamda shu reaksiyalarga tegishli muvozanat doimiylarini muhokama qilamiz: K, start subscript, start text, k, end text, end subscript – kislotaning dissotsilanish doimiysi va K, start subscript, start text, a, end text, end subscript – asosning dissotsilanish doimiysi.

Quyidagi bilan boshlab olamiz: kislota kuchi va start text, p, H, end text ni taqqoslash

1-masala: bir xil konsentratsiyali kuchsiz va kuchli kislotalar

Bizda 2 ta suvli eritma bor: 2, comma, 0, start text, M, end text li ftorid kislota (start text, H, F, end text, left parenthesis, s, point, e, point, right parenthesis) va 2, comma, 0, start text, M, end text bromid kislota (start text, H, B, r, end text, left parenthesis, s, point, e, point, right parenthesis) eritmasi. Qaysi eritmaning start text, p, H, end text qiymati kichikroq?
Bitta javobni tanlang:

2-masala: turli konsentratsiyali kuchli va kuchsiz kislotalar

2, comma, 0, start text, M, end text li ftorid kislota (start text, H, F, end text, left parenthesis, s, point, e, point, right parenthesis) va 1, comma, 0, start text, M, end text li bromid kislota (start text, H, B, r, end text, left parenthesis, s, point, e, point, right parenthesis) eritmalari bor. Qaysi eritmaning start text, p, H, end text qiymati kichikroq?
Aytaylik, ftorid kislota dissotsilanishining muvozanat doimiysini bilmaymiz.
Bitta javobni tanlang:

Kuchsiz kislotalar va K, start subscript, start text, k, end text, end subscript – kislotaning dissotsilanish konstantasi

Eritmada toʻliq dissotsilanmaydigan kislotalar kuchsiz kislotalar hisoblanadi. Boshqacha aytganda, kuchli kislota boʻlmagan har qanday kislota kuchsiz kislotadir.
Kuchsiz kislotaning kuchi uning qanchasi dissotsilanganiga bogʻliq: qanchalik koʻp dissotsilansa, shunchalik kuchli kislota deyish mumkin. Kuchi bir-biriga yaqin boʻlgan kislotalarni oʻzaro taqqoslash uchun kislotaning dissotsilanish reaksiyasi doimiysi boʻlgan K, start subscript, start text, k, end text, end subscript dan foydalanishimiz mumkin.
Bir negizli kuchsiz kislota (start text, H, K, end text) ning dissotsilanish reaksiyasini quyidagicha yozishimiz mumkin:
start text, H, K, end text, left parenthesis, s, point, e, point, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, s, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, s, point, e, point, right parenthesis, plus, start text, K, end text, start superscript, minus, end superscript, left parenthesis, s, point, e, point, right parenthesis
Ushbu reaksiya asosida muvozanat domiysi (K, start subscript, start text, k, end text, end subscript) ifodasini yozishimiz mumkin:
K, start subscript, start text, k, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, K, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, K, end text, close bracket, end fraction
Muvozanat konstantasi bu reaksiya mahsulotlari konsentratsiyasining reagentlar konsentratsiyasiga nisbati. start text, H, K, end text dissotsilanib start text, H, end text, start superscript, plus, end superscript va tutash asos - start text, K, end text, start superscript, minus, end superscript ni hosil qiladi. Kislota qanchalik kuchli boʻlsa K, start subscript, start text, k, end text, end subscript qiymati shunchalik katta boʻladi. K, start subscript, start text, k, end text, end subscript va/yoki kislota konsentratsiyasi ortishi bilan start text, p, H, end text kamayadi.

Asosiy kuchsiz kislotalar

Olma kislota start text, C, end text, start subscript, 4, end subscript, start text, H, end text, start subscript, 6, end subscript, start text, O, end text, start subscript, 5, end subscript olmada uchraydigan organik kislota. Rasm manbasi: Wikimedia Commons, CC BY-SA 3.0.
Karboksil guruhlar organik kislotalarning asosiy guruhlaridan boʻlib, quyidagicha tuzilishga ega: minus, start text, C, O, O, H, end text. Olma kislota left parenthesis, start text, C, end text, start subscript, 4, end subscript, start text, H, end text, start subscript, 6, end subscript, start text, O, end text, start subscript, 5, end subscript, right parenthesis olma va boshqa baʼzi mevalarga nordon taʼm beruvchi, ikkita karboksil guruhli organik kislotadir. Molekulasida ikkita karboksil guruhi borligi uchun olma kislota ikkita proton ajratib chiqaradi.
Quyidagi jadvalda baʼzi kuchsiz kislotalar va ularning K, start subscript, start text, k, end text, end subscript qiymatlari keltirilgan.
NomiFormulasiK, start subscript, start text, k, end text, end subscript, left parenthesis, 25, degrees, start text, C, end text, right parenthesis
Ammoniy ionistart text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript5, comma, 6, times, 10, start superscript, minus, 10, end superscript
Xlorit kislotastart text, H, C, l, O, end text, start subscript, 2, end subscript1, comma, 2, times, 10, start superscript, minus, 2, end superscript
Ftorid kislotastart text, H, F, end text7, comma, 2, times, 10, start superscript, minus, 4, end superscript
Sirka kislotastart text, C, H, end text, start subscript, 3, end subscript, start text, C, O, O, H, end text1, comma, 8, times, 10, start superscript, minus, 5, end superscript
Nazorat savoli: yuqoridagi jadvalga asoslanib qaysi kislota kuchliroq ekanini ayting, sirka yoki ftorid kislota?

1-misol: kuchsiz kislotaning dissotsilanish darajasini hisoblash

Kislota eritmada qanchalik dissotsilanganini koʻrsatadigan yana bir koʻrsatkich dissotsilanish darajasidir. Kuchsiz kislota start text, H, K, end text uchun dissotsilanish darajasi quyidagicha hisoblanadi:
start text, d, i, s, s, o, t, s, i, l, a, n, i, s, h, space, d, a, r, a, j, a, s, i, end text, equals, start fraction, open bracket, start text, K, end text, start superscript, minus, end superscript, left parenthesis, s, point, e, point, right parenthesis, close bracket, divided by, open bracket, start text, H, K, end text, left parenthesis, s, point, e, point, right parenthesis, close bracket, end fraction, times, 100, percent
25, degrees, start text, C, end text da left parenthesis, start text, H, N, O, end text, start subscript, 2, end subscript, right parenthesis ning K, start subscript, start text, k, end text, end subscript qiymati 4, comma, 0, times, 10, start superscript, minus, 4, end superscript ga teng boʻlsa, 0, comma, 400, start text, space, M, end text li eritmada kislotaning dissotsilanish darajasi nechaga teng boʻladi?
Keling, ushbu misolni bosqichma-bosqich koʻrib chiqaylik!

1-qadam: kislotaning tenglashtirilgan dissotsilanish reaksiyasini yozing:

Dastlab start text, H, N, O, end text, start subscript, 2, end subscript ning suvli eritmada dissotsilanish reaksiyasini yozamiz. Nitrit kislota suvga proton ajratib chiqaradi va start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, left parenthesis, s, point, e, point, right parenthesis hosil qiladi:
start text, H, N, O, end text, start subscript, 2, end subscript, left parenthesis, s, point, e, point, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, s, right parenthesis, \rightleftharpoons, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, left parenthesis, s, point, e, point, right parenthesis, plus, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, left parenthesis, s, point, e, point, right parenthesis

2-qadam: K, start subscript, start text, k, end text, end subscript ifodasini yozing

1-qadamdagi tenglamadan foydalanib nitrit kislota uchun K, start subscript, start text, k, end text, end subscript ifodasini yozamiz:
K, start subscript, start text, k, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, N, O, end text, start subscript, 2, end subscript, close bracket, end fraction, equals, 4, comma, 0, times, 10, start superscript, minus, 4, end superscript

3-qadam: muvozanat holatidagi open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket va open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket qiymatlarini toping

Endi start text, B, K, M, end text jadvalidan foydalanib, K, start subscript, start text, k, end text, end subscript ifodamiz uchun algebraik qiymatlar (muvozanat konsentratsiyalari)ni topamiz:
start text, H, N, O, end text, start subscript, 2, end subscript, left parenthesis, s, point, e, point, right parenthesis\rightleftharpoonsstart text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscriptstart text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript
Boshlangʻich0, comma, 400, start text, M, end text00
Kirishganminus, xplus, xplus, x
Muvozanat0, comma, 400, start text, M, end text, minus, xxx
K, start subscript, start text, k, end text, end subscript ifodamizga muvozanat konsentratsiyalarini qoʻyib, quyidagini olamiz:
K, start subscript, start text, k, end text, end subscript, equals, start fraction, left parenthesis, x, right parenthesis, left parenthesis, x, right parenthesis, divided by, left parenthesis, 0, comma, 400, start text, M, end text, minus, x, right parenthesis, end fraction, equals, 4, comma, 0, times, 10, start superscript, minus, 4, end superscript
Yuqoridagi ifodani soddalashtirib, quyidagini olamiz:
start fraction, x, squared, divided by, 0, comma, 400, start text, M, end text, minus, x, end fraction, equals, 4, comma, 0, times, 10, start superscript, minus, 4, end superscript
Bu kvadrat tenglama, undan x ni topish uchun kvadrat formuladan yoki yaqinlashtirish usulidan foydalanishimiz mumkin.
Ikkala usulda ham x, equals, 0, comma, 0126, start text, space, M, end text natijani olamiz. Demak, open bracket, start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript, close bracket, equals, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, 0, comma, 0126, start text, space, M, end text.

4-qadam: dissotsilanish darajasini hisoblash

Dissotsilanish darajasini topish uchun 3-qadamda topilgan muvozanat konsentratsiyalaridan foydalanishimiz mumkin.
Shunday qilib, start text, H, N, O, end text, start subscript, 2, end subscript eritmada 3, comma, 2, percent start text, H, end text, start superscript, plus, end superscript va start text, N, O, end text, start subscript, 2, end subscript, start superscript, minus, end superscript ionlariga ajralgan.

Kuchsiz asos va K, start subscript, start text, a, end text, end subscript

Keling, endi asoslarning dissotsilanish (yoki ionlanish) doimiysi (K, start subscript, start text, a, end text, end subscript)ni koʻrib chiqamiz. Ishni asos (start text, A, end text) uchun umumiy ionlanish reaksiyasini yozishdan boshlaymiz. Ushbu reaksiyada asoslar suvdan proton qabul qilib, gidroksid va tutash kislotasi (start text, A, H, end text, start superscript, plus, end superscript) ni hosil qiladi.
start text, A, end text, left parenthesis, s, point, e, point, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, s, right parenthesis, \rightleftharpoons, start text, A, H, end text, start superscript, plus, end superscript, left parenthesis, s, point, e, point, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, s, point, e, point, right parenthesis
Muvozanat doimiysi (K, start subscript, start text, a, end text, end subscript) uchun quyidagi ifodani yozishimiz mumkin:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, A, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, A, end text, close bracket, end fraction
Ushbu ifodadan koʻrinib turibdiki, asos ionlanganda qancha koʻp start text, A, H, end text, start superscript, plus, end superscript hosil qilsa, shunchalik kuchli hisoblanadi va K, start subscript, start text, a, end text, end subscript qiymati ham shunchalik katta boʻladi. Shunday qilib, start text, p, H, end text koʻrsatkichi K, start subscript, start text, a, end text, end subscript va eritmadagi asos konsentratsiyasiga bogʻliq boʻladi.

2-misol: kuchsiz asos eritmasining start text, p, H, end text ini aniqlash

1, comma, 50, start text, space, M, end text ammiak left parenthesis, start text, N, H, end text, start subscript, 3, end subscript, right parenthesis eritmasining start text, p, H, end text qiymati qanchaga teng? left parenthesis, K, start subscript, start text, a, end text, end subscript, equals, 1, comma, 8, times, 10, start superscript, minus, 5, end superscript, right parenthesis
Ushbu misol bitta qoʻshimcha (start text, p, H, end text dan open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket ni topish) qadami boʻlgan muvozanat masalasidir. Keling, hisob-kitoblarni bosqichma-bosqich koʻrib chiqaylik.

1-qadam: ionlanish tenglamasini yozing

Dastlab ammiak uchun ionlanish reaksiyasini yozamiz. Ammiak suvdan proton qabul qilib, ammoniy ioni left parenthesis, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, right parenthesisni hosil qiladi:
start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, s, point, e, point, right parenthesis, plus, start text, H, end text, start subscript, 2, end subscript, start text, O, end text, left parenthesis, s, right parenthesis, \rightleftharpoons, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, left parenthesis, s, point, e, point, right parenthesis, plus, start text, O, H, end text, start superscript, minus, end superscript, left parenthesis, s, point, e, point, right parenthesis

2-qadam: K, start subscript, start text, a, end text, end subscript ifodasini yozamiz

Yuqoridagi tenglamadan foydalanib, K, start subscript, start text, a, end text, end subscript ifodasini yozishimiz mumkin:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, N, H, end text, start subscript, 3, end subscript, close bracket, end fraction, equals, 1, comma, 8, times, 10, start superscript, minus, 5, end superscript

3-qadam: muvozanat holatidagi open bracket, start text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscript, close bracket va open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket larni toping

Muvozanat konstantalarini aniqlash uchun start text, B, K, M, end text jadvalidan foydalanamiz:
start text, N, H, end text, start subscript, 3, end subscript, left parenthesis, s, point, e, point, right parenthesis\rightleftharpoonsstart text, N, H, end text, start subscript, 4, end subscript, start superscript, plus, end superscriptstart text, O, H, end text, start superscript, minus, end superscript
Boshlangʻich1, comma, 50, start text, M, end text00
Kirishganminus, xplus, xplus, x
Muvozanat1, comma, 50, start text, M, end text, minus, xxx
Muvozanat qiymatlarini K, start subscript, start text, a, end text, end subscript ifodasiga qoʻyib, quyidagini hosil qilamiz:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, left parenthesis, x, right parenthesis, left parenthesis, x, right parenthesis, divided by, 1, comma, 50, start text, M, end text, minus, x, end fraction, equals, 1, comma, 8, times, 10, start superscript, minus, 5, end superscript
Ifodani soddalashtirsak:
start fraction, x, squared, divided by, 1, comma, 50, start text, M, end text, minus, x, end fraction, equals, 1, comma, 8, times, 10, start superscript, minus, 5, end superscript
Bu kvadrat tenglama, uni kvadrat formula yoki yaqinlashtirish usuli bilan yechishimiz mumkin. Ikkala usulda ham tenglamani yechish mumkin
x, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, equals, 5, comma, 2, times, 10, start superscript, minus, 3, end superscript, start text, space, M, end text

4-qadam: start text, p, H, end text dan open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket ni topish

Endi gidroksid konsentratsiyasini bilganimiz uchun start text, p, O, H, end text ni hisoblashimiz mumkin:
pOH=log[OH]=log(5,2×103)=2,28\begin{aligned}\text{pOH}&=-\log[\text{OH}^-]\\ \\ &=-\log(5{,}2\times10^{-3})\\ \\ &=2{,}28\end{aligned}
Eslatma: 25, degrees, start text, C, end text da start text, p, H, end text, plus, start text, p, O, H, end text, equals, 14. Ushbu tenglamani qayta yozib, quyidagi ifodani olamiz:
start text, p, H, end text, equals, 14, minus, start text, p, O, H, end text
start text, p, O, H, end text ifodasiga qiymatlarimizni qoʻyib, quyidagilarni olamiz:
start text, p, H, end text, equals, 14, comma, 00, minus, left parenthesis, 2, comma, 28, right parenthesis, equals, 11, comma, 72
Shunday qilib, eritmaning start text, p, H, end text qiymati 11,72 ekan.

Asosiy kuchsiz asoslar

Chapda piridinning tuzilishi. Oʻngda aminlarning umumiy tuzilishi: neytral azot atomi R1, R2 va R3 bilan uchta oddiy bogʻ hosil qilgan.
Piridin (chapda) azot tutgan siklik (halqali) birikma. Aminlar (oʻngda) vodorod yoki uglerod bilan uchta bogʻ hosil qilgan neytral azot atomi saqlagan organik birikmalardir. Ikkala molekula ham asos xossasini namoyon qiladi.
Sovunlardan tortib maishiy tozalagichlargacha atrofimiz kuchsiz asoslar bilan oʻralgan. Aminlar – boshqa atomlar (odatda uglerod yoki vodorod) bilan uchta bogʻ hosil qilgan neytral azot atomlari kuchsiz organik asoslardagi asosiy funksional guruhlar hisoblanadi.
Azotning erkin elektron jufti proton qabul qila olgani uchun aminlar asos xossasini namoyon qila oladi. Ammiak left parenthesis, start text, N, H, end text, start subscript, 3, end subscript, right parenthesis amin asoslariga misol boʻla oladi. Azot saqlagan asoslardan yana biri piridin left parenthesis, start text, C, end text, start subscript, 5, end subscript, start text, H, end text, start subscript, 5, end subscript, start text, N, end text, right parenthesisdir.

Xulosa

  • Bir negizli, kuchsiz kislotaning start text, H, K, end text tutash asosi bilan umumiy muvozanat konstantasi quyidagicha boʻladi:
K, start subscript, start text, k, end text, end subscript, equals, start fraction, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, K, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, H, K, end text, close bracket, end fraction
  • K, start subscript, start text, k, end text, end subscript – kislotaning dissotsilanish doimiysi kuchsiz kislota dissotsilanishining miqdoriy oʻlchovini koʻrsatuvchi kattalik hisoblanadi. K, start subscript, start text, k, end text, end subscript qiymati qanchalik katta boʻlsa, kislota shunchalik kuchli boʻladi va aksincha.
  • Kuchsiz asosning tutash kislotasi bilan muvozanat doimiysi ifodasi quyidagicha boʻladi:
K, start subscript, start text, a, end text, end subscript, equals, start fraction, open bracket, start text, A, H, end text, start superscript, plus, end superscript, close bracket, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, divided by, open bracket, start text, A, end text, close bracket, end fraction
  • Asosning dissotsilanish doimiysi (yoki asosning ionlanish doimiysi) K, start subscript, start text, a, end text, end subscript kuchsiz asosning ionlanish darajasini koʻrsatadi. K, start subscript, start text, a, end text, end subscript qanchalik katta boʻlsa, asos shunchalik kuchli boʻladi va aksincha.

Urinib koʻring!

1-masala: start text, p, H, end text dan K, start subscript, start text, a, end text, end subscript ni topish

25, degrees, start text, C, end text da 1, comma, 50, start text, space, M, end text li piridin left parenthesis, start text, C, end text, start subscript, 5, end subscript, start text, H, end text, start subscript, 5, end subscript, start text, N, end text, right parenthesis eritmasining start text, p, H, end text i 9, comma, 70 ga teng. Piridin uchun K, start subscript, start text, a, end text, end subscript ni toping.
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