Question

For a titration of 10.0 mL of 0.277 M NaOH, what would be the pH after addition of 32.7 mL of 0.13 M HNO3?

Answer 1

we have:

Molarity of HNO3 = 0.13 M

Volume of HNO3 = 32.7 mL

Molarity of NaOH = 0.277 M

Volume of NaOH = 10 mL

mol of HNO3 = Molarity of HNO3 * Volume of HNO3

mol of HNO3 = 0.13 M * 32.7 mL = 4.251 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.277 M * 10 mL = 2.77 mmol

We have:

mol(HNO3) = 4.251 mmol

mol(NaOH) = 2.77 mmol

2.77 mmol of both will react

remaining mol of HNO3 = 1.481 mmol

Total volume = 42.7 mL

[H+]= mol of acid remaining / volume

[H+] = 1.481 mmol/42.7 mL

= 0.0347 M

we have below equation to be used:

pH = -log [H+]

= -log (3.468*10^-2)

= 1.46

Answer: 1.46