Question

What is the pH of 70 mL of 0.01 M HCl solution after 30 mL of 0.0075 M NaOH have been added? What is the pH at the equivalence point?

Answer 1

1)
Given:
M(HCl) = 0.1 M
V(HCl) = 70 mL
M(NaOH) = 0.0075 M
V(NaOH) = 30 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.1 M * 70 mL = 7 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.0075 M * 30 mL = 0.225 mmol


We have:
mol(HCl) = 7 mmol
mol(NaOH) = 0.225 mmol
0.225 mmol of both will react
remaining mol of HCl = 6.775 mmol
Total volume = 100.0 mL

[H+]= mol of acid remaining / volume
[H+] = 6.775 mmol/100.0 mL
= 0.0678 M


use:
pH = -log [H+]
= -log (6.775*10^-2)
= 1.17

Answer: 1.17

2)
Since this is strong acid and strong base reaction
pH at equivalence point will be 7

Answer: 7.0