Write the chemical equations for the dissociation of each proton in phosphoric a
ID: 1039152 • Letter: W
Question
Write the chemical equations for the dissociation of each proton in phosphoric acid, H,POs Label all conjugate acid-base pairs and indicate which of the equations would have the smallest K, value. .Calculate (H 0'] and the pH of a 0.02M solution of Sr(OH)2 Calculate [H 0'] and the pH of a 0.80 M solution of H3PO. K, 6.9 x 103. Calculate [H3O'] and the pH of a 0.02M solution of fluoride ion, F. The K, of hydrofluoric acid, HF is 6.6 x 104. (Find Ko first then work out the problem like a weak base)Explanation / Answer
a. H3PO4 ----------------> H^+ (aq) + H2PO4^- (aq) Ka1
H2PO4^- (aq) ----------> HPO4^2- (aq) + H^+ (aq) Ka2
HPO4^2- (aq) -----------> H^+ (aq) + PO4^3- (aq) Ka3 least acidic
Ka3 is smallest value
Sr(OH)2(aq) --------------> Sr^2+ (aq) + 2OH^- (aq)
0.02M 2*0.02M
[OH^-] = 2*[Sr(OH)2]
[OH^-] = 2*0.02M
[OH^-] = 0.04M
[H3O^+] = Kw/[OH^-]
= 1*10^-14/0.04 = 2.5*10^-13M
PH = -log[H3O^+]
= -log2.5*10^-13
= 12.6
H3PO4(aq) -------------> H^+ (aq) + H2PO4^- (aq)
I 0.8 0 0
C -x +x +x
E 0.8-x +x +x
Ka1 = [H^+][H2PO4^-]/[H3PO4]
6.9*10^-3 = x*x/0.8-x
6.9*10^-3 *(0.8-x) = x^2
x = 0.071
[H^+] = x = 0.071M
PH = -log[H^+]
= -log0.071
= 1.1487
F^- + H2O ------------------> HF + OH^-
I 0.02 0 0
C -x +x +x
E 0.02-x +x +x
Kb = Kw/Ka
= 1*10^-14/6.6*10^-4 = 1.5*10^-11
Kb = [HF][OH^-]/[F^-]
1.5*10^-11 = x*x/0.02-x
1.5*10^-11*(0.02-x) = x^2
x = 5.47*10^-7
[OH^-] = x = 5.47*10^-7 M
[H3O^+] = 1*10^-14/5.47*10^-7 = 1.83*10^-8 M
PH = -log[H3O^+]
= -log1.83*10^-8
= -log1.83*10^-8
= 7.7375
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