Calculate the p H of the solution that results from each of the following mixtur

Question

Calculate the pH of the solution that results from each of the following mixtures. Part A 140.0mL of 0.27M HF with 230.0mL of 0.31M NaF Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Part B 185.0mL of 0.11M C2H5NH2 with 280.0mL of 0.21M C2H5NH3Cl Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Calculate the pH of the solution that results from each of the following mixtures. Part A 140.0mL of 0.27M HF with 230.0mL of 0.31M NaF Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Calculate the pH of the solution that results from each of the following mixtures. Part A 140.0mL of 0.27M HF with 230.0mL of 0.31M NaF Express your answer using two decimal places. pH = SubmitMy AnswersGive Up pH = pH = SubmitMy AnswersGive Up Part B 185.0mL of 0.11M C2H5NH2 with 280.0mL of 0.21M C2H5NH3Cl Express your answer using two decimal places. pH = SubmitMy AnswersGive Up Part B 185.0mL of 0.11M C2H5NH2 with 280.0mL of 0.21M C2H5NH3Cl Express your answer using two decimal places. pH = SubmitMy AnswersGive Up pH = pH = SubmitMy AnswersGive Up pH =

Explanation / Answer

A)

we need to recalculate molarities after we combine the solutions:
.27 M = x moles HF / .140 L
We have 0.0378 moles HF
.31M = x moles NaF / .230 L
We have 0.0713 moles NaF
Now, there are 0.140 L + 0.230 L = 0.370 L of solution, so we have:
0.0378 moles HF / 0.370 L = 0.1022 M HF
0.0713 moles NaF / 0.370 L = 0.1927M NaF

Okay, so we need to determine all the concentrations in solution: we know that HF is a weak acid, so it will dissociate to some extent according to its acid dissociation constant, or Ka. NaF completely dissociates into Na+ and F-.
So, to begin with, we have .1927 M Na+ and .1927 M F-. Now, we can figure out how much of the HF dissociates, so we can figure out how much H+ is in solution, which will directly lead us to pH.

Ka for HF = 7.2*10^-4
Ka = [H+][F-] / [HF]
where brackets indicate concentration. So, after some amount x dissociates, we can write this expression:

7.2*10^-4 = (x)(.1927+ x) / (.1022 - x)
Where there's already .196 F- in solution. Now, we can approximate x to be negligibly small, so now we can write:

7.2*10^-4 = (x)(.1927) / (.1022)
x = 0.000382 M=[ H+]

pH = -log[H+]
pH = -log(0.000382)
pH = 3.418

B)

Kb of C2H5NH2=5.6*10^-4,

5.6*10^-4=x[280*0.21/ 185*0.11]=x(2.889)

x=[OH-]=1.938*10^-4M

pOH=3.7126

pH=14-pOH=10.287

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