Question

Find the pH of each of the following solutions of mixtures of acids.​

Part A

0.190 M in HCHO2 and 0.23 M in HC2H3O2​

Part B

6.0×10⁻² M in acetic acid and 6.0×10⁻² M in hydrocyanic acid

Answer 1

I have a readymade for you so please replace only the two concentrations of HCO2 and HC2H3O2 by 1.80M and 0.215M. and do practice for other concentrations.

0.180 M in HCHO₂ and 0.215 M in HC₂H₃O₂
The Ka of HCHO₂ is 1.8 x 10⁻⁴ and the Ka of HC₂H₃O₂ is 1.8 x 10⁻⁵
Let's start with HCHO₂
___HCHO₂ + H₂O <----------> CHO₂⁻ + H₃O⁺
I___0.180M_______________0M_____0M
C___ - x_________________+ x______+ x
E__0.180 - x______________x________x

Ka = [CHO₂⁻] * [H₃O⁺] / [HCHO₂]
1.8 x 10⁻⁴ = [x] * [x] / [0.180 - x]
1.8 x 10⁻⁴ = x² / (0.180 - x)
Drop x from the denominator so that (0.180 - x) becomes (0.180)
1.8 x 10⁻⁴ = x² / (0.180)
(1.8 x 10⁻⁴) *(0.180)= x²
3.24 x 10⁻⁵ = x²
√(3.24 x 10⁻⁵) = x
.0057 = x
If you use a quadratic equation you get .0056M for x, but that little difference is okay.
Then set up the ICE chart for HC₂H₃O₂ but instead of 0 for the initial H₃O⁺, put .0057
___HC₂H₃O₂ + H₂O <----------> C₂H₃O₂⁻ + H₃O⁺
I___0.215M_______________0M_____.0057M
C___ - x_________________+ x______+ x
E__0.215 - x______________x________.0057 + x

Ka = [C₂H₃O₂⁻ ] * [H₃O⁺] / HC₂H₃O₂]
1.8 x 10⁻⁵ = [x] * [.0057 + x] / [0.215 - x]
Drop the x in both [.0057 + x] and [0.215 - x], so that it becomes [.0057] and [.215]
1.8 x 10⁻⁵ = [x] * [.0057] / [0.215]
(1.8 x 10⁻⁵) * (.215) / (.0057) = [x]
6.8 x 10⁻⁴ = x

Since [H₃O⁺] is [.0057 + x] and x is 6.8 x 10⁻⁴, [H₃O⁺] is
.0057 + 6.8 x 10⁻⁴ = .0064
pH = -log [H₃O⁺] = -log [.0064] = 2.19
If you use quadratics instead, your answer will be like 2.21, but the pH i calculated is really close and simpler without all the quadratic.

For ka values see http://chemistry.about.com/od/chartstabl...

PartB) In the same way you please do for this part also.