Sodium fluoride is added to many municipal water supplies to reduce tooth decay. Calculate the pH of a 3.12×10-3 M solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.
Solution
NaF(aq) --------------> Na^+ (aq) + F^- (aq)
3.12*10^-3 3.12*10^-3M
F^- (aq) + H2O -------------> HF (aq) + OH^- (aq)
I 0.00312 0 0
C -x +x +x
E 0.00312-x +x +x
Kb = Kw/Ka
= 1*10^-14/6.8*10^-4 = 1.47*10^-11
Kb = [HF][OH^-]/[F^-]
1.47*10^-11 = x*x/0.00312-x
1.47*10^-11 *(0.00312-x) = x^2
x = 2.17*10^-7
[OH^-] = x = 2.17*10^-7 M
POH = -log[OH^-]
= -log2.17*10^-7
= 6.6635
PH = 14-POH
= 14-6.6635
= 7.3365
.
Sodium fluoride is added to many municipal water supplies to reduce to.docx
1. Sodium fluoride is added to many municipal water supplies to reduce tooth decay. Calculate the
pH of a 3.12×10-3 M solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.
Solution
NaF(aq) --------------> Na^+ (aq) + F^- (aq)
3.12*10^-3 3.12*10^-3M
F^- (aq) + H2O -------------> HF (aq) + OH^- (aq)
I 0.00312 0 0
C -x +x +x
E 0.00312-x +x +x
Kb = Kw/Ka
= 1*10^-14/6.8*10^-4 = 1.47*10^-11
Kb = [HF][OH^-]/[F^-]
1.47*10^-11 = x*x/0.00312-x
1.47*10^-11 *(0.00312-x) = x^2
x = 2.17*10^-7
[OH^-] = x = 2.17*10^-7 M
POH = -log[OH^-]
= -log2.17*10^-7
= 6.6635
![Sodium fluoride is added to many municipal water supplies to reduce tooth decay. Calculate the
pH of a 3.12×10-3 M solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.
Solution
NaF(aq) --------------> Na^+ (aq) + F^- (aq)
3.12*10^-3 3.12*10^-3M
F^- (aq) + H2O -------------> HF (aq) + OH^- (aq)
I 0.00312 0 0
C -x +x +x
E 0.00312-x +x +x
Kb = Kw/Ka
= 1*10^-14/6.8*10^-4 = 1.47*10^-11
Kb = [HF][OH^-]/[F^-]
1.47*10^-11 = x*x/0.00312-x
1.47*10^-11 *(0.00312-x) = x^2
x = 2.17*10^-7
[OH^-] = x = 2.17*10^-7 M
POH = -log[OH^-]
= -log2.17*10^-7
= 6.6635](https://image.slidesharecdn.com/sodiumfluorideisaddedtomanymunicipalwatersuppliestoreduceto-230201192450-da594584/85/sodium-fluoride-is-added-to-many-municipal-water-supplies-to-reduce-todocx-1-320.jpg)
