Question

Estimating pH 1. If you combine 40.0 mL of a 0.80 M HF solution with 60.0 mL of a 0.60 M NaF solution, which of the following is correct? HF pKa = 3.17. a) pH will be > 3.17 b) pH will be < 3.17 c) pH will equal 3.17 2. If 1.50 mL of 1.0 M HNO3 is added to a solution containing 40.0 mL of 0.80 M HF and 60.0 mL of 0.60 M NaF, what will happen to the pH of the solution? HF pKa = 3.17. a) pH will increase b) pH will decrease c) pH will be the same as it was before.

Answer 1

1.

Number of moles of HF in 40.0 mL of 0.80 M solution is

  

0.80 mol 1L 1L x 40.0 mL = 0.032 mol HF 1000 mL

Number of moles of NaF in 60.0 mL of 0.60 M solution is

0.60 mol 11 - X 60.0 mL = 0.036 mol NaF 1 L - 1000 mL

NaF dissociates completely as follows:

NaF → Na+ + F-

Hence, 0.036 mol of NaF gives 0.036 mol of F-.

Hence, at equilibrium our acid base mixture contains 0.032 mol of HF and 0.036 mol.

Total volume of the solution = 40.0 mL + 60.0 mL = 100.0 mL = 0.100 L

Hence, the equilibrium concentrations of different species in solution is

F = 0.036 mol -= 0.360 M 0.100 L

HF] = 0.032 mol -= 0.320 M 0.100 L

Hence, we can write the following expression for Ka of HF

HF = H+ + F

K. (H+][F] Ka= [H+] x 0.36 M 0.32 M HF

Taking negative logarithm on both sides

Ka- fH+1F-1 (H+x 0.36 M [HF] 0.32 M H+] x 0.36 M - log Ka = – log 0.32 M 0.36 pKq = -log[H+] - log 3.17 = pH - 0.05 pH = 3.17 + 0.05 = 3.22

Hence, the pH of the solution will be greater than 3.17.

Hence, the correct choice is a) pH will be > 3.17

2.

Addition of HNO3 to the above solution will result in production of HF and consuming of F- as follows:

HNO3(aq) + Naf NaNO3 + HF

The number of moles of HNO3 i.e number of moles of H+ added by addition of 1.50 mL of 1.0 M HNO3 is

1.0 mol 1 L x 1.50 mL = 0.0015 mol H+ * 1000 ml

Hence, we can create the following ICE for the addition of H+ by HNO3 in the reaction

	$H^+$	$F^-$	+ HF
Initial, mol	0	0.036	0.032
Change, mol	+0.0015	-0.0015	+0.0015
Equilibrium, mol	0	0.0345	0.0335

Note that since volume is common for all, the ratio of moles is equal to ratio of concentrations.

Now, using the Ka expression of HF

Ka- fH+1F-1 (H+ x 0.345 M [HF] 0.335 M (H+] x 0.345 M = -log Ka = – log ( 0.335 M 0.345 →pka = -log[H+] - log99 3.17 = pH – 0.012 pH = 3.17 + 0.012 = 3.182

Note that pH of the solution was 3.22 before addition of HNO₃.

Now it has decreased to 3.182.

Hence, the correct choice is b) pH will decrease.