Calculate equilibrium concentrations for [H], [H2 PO4-,], [H3PO4] ;, [HPO4-2]; [PO4-3] if at the beginning the...

Calculate equilibrium concentrations for [H], [H2 PO4-,], [H3PO4] ;, [HPO4-2]; [PO4-3] if at the beginning the concentration of [H3PO4 is 0.100 M if K1=7.5 x 10^-3 for the first dissociation, and K 2= 6.2 x 10 ^-8, K3= 4.8 x10^-8

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Expert Solution

-----H3Po4(aq) ------------> H^+ (aq) + H2PO4^- (aq)

I---- 0.1 --------------------- 0 ---------------- 0

C--- -x --------------------- +x --------------- +x

E ---- 0.1-x --------------- +x ----------------- +x

ka1 = [H^+][H2PO4^-]/[H3Po4]

7.5*10^-3 = x*x/(0.1-x)

7.5*10^-3 *(0.1-x) = x^2

x = 0.024M

[H^+] = x = 0.024M

[H2PO4^-] = x = 0.024M

[H3PO4] = 0.1-x = 0.1-0.024 = 0.076M

H2PO4^- (aq) -----------> H^+ (aq) + HPO4^2- (aq)

Ka2 = [H^+][HPO4^2-]/[H2Po4^-]

6.2*10^-8 = 0.024*[HPO4^2-]/0.024

[HPO4^2-] = 6.2*10^-8M

HPO4^2- (aq)--------------> H^+ (aq) + PO4^3- (aq)

Ka3 = [H^+][PO4^3-]/[HPO4^2-]

4.8*10^-8 = 0.024[PO4^3-]/6.2*10^-8

[PO4^3-] = 4.8*10^-8*6.2*10^-8/0.024 = 1.24*10^-13M

[H^+] = 0.024M

[H2PO4^-] = 0.024M

[H3PO4] = 0.076M

[HPO4^2-] = 6.2*10^-8M

[PO4^3-] = 1.24*10^-13M

queen_honey_blossom answered 1 year ago

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