1 Calculate the pH expected when 10 mL of 010 M NaOH is adde

1. Calculate the pH expected when 10 mL of 0.10 M NaOH is added to 20 mL of water (hint: dilution).

2. Calculate the pH expected when 10 mL of 0.10 M NaOH is added to 20 mL of 0.10 M HF (Ka of HF = 7.1 x 10-4 ).

Solution

1)
use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 0.1 M
V1 = 10.0 mL
V2 = 30.0 mL

use:
M1*V1 = M2*V2
M2 = (M1*V1)/V2
M2 = (0.1*10)/30
M2 = 0.0333 M
This is concentration of NaOH after dilution

use:
pOH = -log [OH-]
= -log (3.33*10^-2)
= 1.4776


use:
PH = 14 - pOH
= 14 - 1.4776
= 12.52
Answer: 12.52

2)
Given:
M(HF) = 0.1 M
V(HF) = 20 mL
M(NaOH) = 0.1 M
V(NaOH) = 10 mL


mol(HF) = M(HF) * V(HF)
mol(HF) = 0.1 M * 20 mL = 2 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 10 mL = 1 mmol


We have:
mol(HF) = 2 mmol
mol(NaOH) = 1 mmol

1 mmol of both will react

excess HF remaining = 1 mmol
Volume of Solution = 20 + 10 = 30 mL
[HF] = 1 mmol/30 mL = 0.0333M

[F-] = 1/30 = 0.0333M

They form acidic buffer
acid is HF
conjugate base is F-


Ka = 7.1*10^-4

pKa = - log (Ka)
= - log(7.1*10^-4)
= 3.149

use:
pH = pKa + log {[conjugate base]/[acid]}
= 3.149+ log {3.333*10^-2/3.333*10^-2}
= 3.149


Answer: 3.15