MBCB Larutan Natrium Dihidrogen Fosfat

Diskusi bidang kimia

Moderators: HeadAdmin, CrownedP

Post Reply
masbambang
Member
Member
Posts: 23
Joined: Thu Oct 04, 2018 7:01 pm
Bidang OSN: Kimia SMA
Tahun OSN: 2018
Deskripsi: mantap gan :D
Has thanked: 4 times
Been thanked: 3 times

Tue Mar 05, 2019 11:09 pm

$\ce{ \begin{align*} H3PO4 ${(aq)}$ &<=> H+ ${(aq)}$ + H2PO4- ${(aq)}$ &K_{a \space 1} &= 7.1*10^{-3} \\ H2PO4- ${(aq)}$ &<=> H+ ${(aq)}$ + HPO4^{2-} ${(aq)}$ &K_{a \space 2} &= 6.3*10^{-8} \\ HPO4^{2-} ${(aq)}$ &<=> H+ ${(aq)}$ + PO4^{3-} ${(aq)}$ &K_{a \space 3} &= 4.5*10^{-13} \end{align*} }$
  1. Asumsi 1: Semua spesi $\ce{ H3PO4 }$ terbentuk
    $\ce{ \begin{align*} C_{0 \space H_2PO_4^-} &= [H3PO4] + [H2PO4-] + [HPO4^{2-}] + [PO4^{3-}] \\ &= [H2PO4-]\left(1 + \frac{[H+]}{K_{a \space 1}} + \frac{K_{a \space 2}}{[H+]} + \frac{K_{a \space 2}K_{a \space 3}}{[H+]^2}\right) \\ &= [H2PO4-]\left(\frac{[H+]^3 + K_{a \space 1}[H+]^2 + K_{a \space 1}K_{a \space 2}[H+] + K_{a \space 1}K_{a \space 2}K_{a \space 3}}{K_{a \space 1}[H+]^2}\right) \\ [H2PO4-] &= \frac{K_{a \space 1}C_{0 \space H_2PO_4^-}[H+]^2}{[H+]^3 + K_{a \space 1}[H+]^2 + K_{a \space 1}K_{a \space 2}[H+] + K_{a \space 1}K_{a \space 2}K_{a \space 3}} \end{align*} }$

    $\ce{ \begin{align*} [Na+] &= [OH-] + [H2PO4-] + 2[HPO4^{2-}] + 3[PO4^{3-}] \\ &= \frac{K_w}{[H+]} + \frac{K_{a \space 1}C_{0 \space H_2PO_4^-}[H+]^2 + 2K_{a \space 1}K_{a \space 2}C_{0 \space H_2PO_4^-}[H+] + 3K_{a \space 1}K_{a \space 2}K_{a \space 3}C_{0 \space H_2PO_4^-}}{[H+]^3 + K_{a \space 1}[H+]^2 + K_{a \space 1}K_{a \space 2}[H+] + K_{a \space 1}K_{a \space 2}K_{a \space 3}} \\ [Na+] &= \frac{[H+]^3(K_w + K_{a \space 1}C_{0 \space H_2PO_4^-}) + K_{a \space 1}[H+]^2(K_w + 2K_{a \space 2}C_{0 \space H_2PO_4^-}) + K_{a \space 1}K_{a \space 2}[H+](K_w + 3K_{a \space 3}C_{0 \space H_2PO_4^-}) + K_wK_{a \space 1}K_{a \space 2}K_{a \space 3}}{[H+]^4 + K_{a \space 1}[H+]^3 + K_{a \space 1}K_{a \space 2}[H+]^2 + K_{a \space 1}K_{a \space 2}K_{a \space 3}[H+]} \\ 0 &= [Na+][H+]^4 + (K_{a \space 1}([Na+] - C_{0 \space H_2PO_4^-}) - K_w)[H+]^3 + K_{a \space 1}(K_{a \space 2}([Na+] - 2C_{0 \space H_2PO_4^-}) - K_w)[H+]^2 + K_{a \space 1}K_{a \space 2}(K_{a \space 3}([Na+] - 3C_{0 \space H_2PO_4^-}) - K_w)[H+] - K_wK_{a \space 1}K_{a \space 2}K_{a \space 3}\end{align*} }$

    $\ce{ [Na+] = C_{0 \space H_2PO_4^-} = [NaH2PO4] }$, maka
    $\ce{ 0 = [NaH2PO4][H+]^4 - K_w[H+]^3 - K_{a \space 1}([NaH2PO4] + K_w)[H+]^2 - K_{a \space 1}K_{a \space 2}(2[NaH2PO4] + K_w)[H+] - K_wK_{a \space 1}K_{a \space 2}K_{a \space 3} }$
  2. Asumsi 2: Tidak terbentuk spesi $\ce{ H3PO4 }$
    $\ce{ \begin{align*} C_{0 \space H_2PO_4^-} &= [H2PO4-] + [HPO4^{2-}] + [PO4^{3-}] \\ &= [H2PO4-]\left(1 + \frac{K_{a \space 2}}{[H+]} + \frac{K_{a \space 2}K_{a \space 3}}{[H+]^2}\right) \\ C_{0 \space H_2PO_4^-} &= [H2PO4-]\left(\frac{[H+]^2 + K_{a \space 2}[H+] + K_{a \space 2}K_{a \space 3}}{[H+]^2}\right) \\ [H2PO4-] &= \frac{[H+]^2C_{0 \space H_2PO_4^-}}{[H+]^2 + K_{a \space 2}[H+] + K_{a \space 2}K_{a \space 3}}\end{align*} }$

    $\ce{ \begin{align*} [Na+] &= [OH-] + [H2PO4-] + 2[HPO4^{2-}] + 3[PO4^{3-}] \\ &= \frac{K_w}{[H+]} + \frac{[H+]^2C_{0 \space H_2PO_4^-} + 2K_{a \space 2}[H+]C_{0 \space H_2PO_4^-} + 3K_{a \space 2}K_{a \space 3}C_{0 \space H_2PO_4^-}}{[H+]^2 + K_{a \space 2}[H+] + K_{a \space 2}K_{a \space 3}} \\ [Na+] &= \frac{[H+]^3C_{0 \space H_2PO_4^-} + [H+]^2(2K_{a \space 2}C_{0 \space H_2PO_4^-} + K_w) + K_{a \space 2}[H+](3K_{a \space 3}C_{0 \space H_2PO_4^-} + K_w) + K_wK_{a \space 2}K_{a \space 3}}{[H+]^3 + K_{a \space 2}[H+]^2 + K_{a \space 2}K_{a \space 3}[H+]} \\ 0 &= ([Na+] - C_{0 \space H_2PO_4^-})[H+]^3 + (K_{a \space 2}([Na+] - 2C_{0 \space H_2PO_4^-}) - K_w)[H+]^2 + K_{a \space 2}(K_{a \space 3}([Na+] - 3C_{0 \space H_2PO_4^-}) - K_w)[H+] - K_wK_{a \space 2}K_{a \space 3}\end{align*} }$

    $\ce{ [Na+] = C_{0 \space H_2PO_4^-} = [NaH2PO4] }$, maka
    $\ce{ 0 = (K_{a \space 2}[NaH2PO4] + K_w)[H+]^2 + K_{a \space 2}(2K_{a \space 3}[NaH2PO4] + K_w)[H+] + K_wK_{a \space 2}K_{a \space 3} }$
    karena akar-akarnya bernilai negatif, maka asumsi ini INVALID
  3. Asumsi 3: Hanya muncul spesi $\ce{ H3PO4 }$ dan $\ce{ H2PO4- }$ pada kesetimbangan
    $\ce{ \begin{align*} C_{0 \space H_2PO_4^-} &= [H3PO4] + [H2PO4-] \\ &= [H2PO4-]\left(1 + \frac{[H+]}{K_{a \space 1}}\right) \\ C_{0 \space H_2PO_4^-} &= [H2PO4-]\left(\frac{[H+] + K_{a \space 1}}{K_{a \space 1}}\right) \\ [H2PO4-] &= \frac{K_{a \space 1}C_{0 \space H_2PO_4^-}}{[H+] + K_{a \space 1}}\end{align*} }$

    $\ce{ \begin{align*} [Na+] &= [OH-] + [H2PO4-] \\ &= \frac{K_w}{[H+]} + \frac{K_{a \space 1}C_{0 \space H_2PO_4^-}}{[H+] + K_{a \space 1}} \\ [Na+] &= \frac{K_wK_{a \space 1} + (K_w + K_{a \space 1}C_{0 \space H_2PO_4^-})[H+]}{[H+]^2 + K_{a \space 1}[H+]} \\ 0 &= [Na+][H+]^2 + (K_{a \space 1}([Na+] - C_{0 \space H_2PO_4^-}) - K_w)[H+] - K_wK_{a \space 1}\end{align*} }$
    $\ce{ [Na+] = C_{0 \space H_2PO_4^-} = [NaH2PO4] }$, maka
    $\ce{ \begin{align*} 0 &= [NaH2PO4][H+]^2 - K_w[H+] - K_wK_{a \space 1} \\ [H+] &= \frac{K_w \pm \sqrt{K_w^2 + 4K_wK_{a \space 1}[NaH_2PO_4]}}{2[NaH2PO4]} \\ \therefore {pH} &= log\left(\frac{2[NaH2PO4]}{K_w \pm \sqrt{K_w^2 + 4K_wK_{a \space 1}[NaH_2PO_4]}}\right)\end{align*} }$
  4. Asumsi 4: Tidak terbentuk spesi $\ce{ PO4^{3-} }$
    $\ce{ \begin{align*} C_{0 \space H_2PO_4^-} &= [H3PO4] + [H2PO4-] + [HPO4^{2-}] \\ &= [H2PO4-]\left(1 + \frac{[H+]}{K_{a \space 1}} + \frac{K_{a \space 2}}{[H+]}\right) \\ C_{0 \space H_2PO_4^-} &= [H2PO4-]\left(\frac{[H+]^2 + K_{a \space 1}[H+] + K_{a \space 1}K_{a \space 2}}{K_{a \space 1}[H+]}\right) \\ [H2PO4-] &= \frac{K_{a \space 1}C_{0 \space H_2PO_4^-}[H+]}{[H+]^2 + K_{a \space 1}[H+] + K_{a \space 1}K_{a \space 2}}\end{align*} }$

    $\ce{ \begin{align*} [Na+] &= [OH-] + [H2PO4-] + 2[HPO4^{2-}] \\ &= \frac{K_w}{[H+]} + \frac{K_{a \space 1}C_{0 \space H_2PO_4^-}[H+] + 2K_{a \space 1}K_{a \space 2}C_{0 \space H_2PO_4^-}}{[H+]^2 + K_{a \space 1}[H+] + K_{a \space 1}K_{a \space 2}} \\ [Na+] &= \frac{[H+]^2(K_{a \space 1}C_{0 \space H_2PO_4^-} + K_w) + K_{a \space 1}[H+](2K_{a \space 2}C_{0 \space H_2PO_4^-} + K_w) + K_wK_{a \space 1}K_{a \space 2}}{[H+]^3 + K_{a \space 1}[H+]^2 + K_{a \space 1}K_{a \space 2}[H+]} \\ 0 &= [Na+][H+]^3 + [H+]^2(K_{a \space 1}([Na+] - C_{0 \space H_2PO_4^-}) - K_w) + K_{a \space 1}[H+](K_{a \space 2}([Na+] - 2C_{0 \space H_2PO_4^-}) - K_w) - K_wK_{a \space 1}K_{a \space 2}\end{align*} }$
    $\ce{ [Na+] = C_{0 \space H_2PO_4^-} = [NaH2PO4] }$, maka
    $\ce{ 0 = [NaH2PO4][H+]^3 - K_w[H+]^2 - K_{a \space 1}(K_{a \space 2}[NaH2PO4] + K_w)[H+] - K_wK_{a \space 1}K_{a \space 2} }$
  5. Asumsi 5: Hanya terbentuk spesi $\ce{ H2PO4- }$ dan $\ce{ HPO4^{2-} }$ pada kesetimbangan
    $\ce{ \begin{align*} C_{0 \space H_2PO_4^-} &= [H2PO4-] + [HPO4^{2-}] \\ &= [H2PO4-]\left(1 + \frac{K_{a \space 2}}{[H+]}\right) \\ &= [H2PO4-]\left(\frac{[H+] + K_{a \space 2}}{[H+]}\right) \\ [H2PO4-] &= \frac{[H+]C_{0 \space H_2PO_4^-}}{[H+] + K_{a \space 2}}\end{align*} }$

    $\ce{ \begin{align*} [Na+] &= [OH-] + [H2PO4-] + 2[HPO4^{2-}] \\ &= \frac{K_w}{[H+]} + \frac{[H+]C_{0 \space H_2PO_4^-} + 2K_{a \space 2}C_{0 \space H_2PO_4^-}}{[H+] + K_{a \space 2}} \\ [Na+] &= \frac{[H+]^2C_{0 \space H_2PO_4^-} + [H+](2K_{a \space 2}C_{0 \space H_2PO_4^-} + K_w) + K_wK_{a \space 2}}{[H+]^2 + K_{a \space 2}[H+]} \\ 0 &= ([Na+] - C_{0 \space H_2PO_4^-})[H+]^2 + (K_{a \space 2}([Na+] - 2C_{0 \space H_2PO_4^-}) - K_w)[H+] - K_wK_{a \space 2}\end{align*} }$
    $\ce{ [Na+] = C_{0 \space H_2PO_4^-} = [NaH2PO4] }$, maka
    $\ce{ \begin{align*} 0 &= - (K_{a \space 2}[NaH2PO4] + K_w)[H+] - K_wK_{a \space 2} \\ [H+] &= -\frac{K_wK_{a \space 2}}{K_{a \space 2}[NaH2PO4] + K_w}\end{align*} }$
    karena $\ce{ [H+] }$ bernilai negatif, maka asumsi ini INVALID
Post Reply