Question

A 20.00 mL sample of 0.120 M NaOH was titrated with 52.00 mL of 0.175 M HNO₃.  

          Calculate the [H⁺], [OH^-], pH, and pOH for the resulting solution.   

Answer 1

Given:

M(HCl) = 0.175 M

V(HCl) = 52 mL

M(NaOH) = 0.12 M

V(NaOH) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.175 M * 52 mL = 9.1 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.12 M * 20 mL = 2.4 mmol

We have:

mol(HCl) = 9.1 mmol

mol(NaOH) = 2.4 mmol

2.4 mmol of both will react

remaining mol of HCl = 6.7 mmol

Total volume = 72.0 mL

[H+]= mol of acid remaining / volume

[H+] = 6.7 mmol/72.0 mL

= 9.306*10^-2 M

use:

pH = -log [H+]

= -log (9.306*10^-2)

= 1.03

Now use:

pOH = 14 - pH

= 14 - 1.03

= 12.97

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(9.306*10^-2)

[OH-] = 1.075*10^-13 M

[H+] = 9.31*10^-2 M

[OH-] = 1.07*10^-13 M

pH = 1.03

pOH = 12.97