Question

19. If 25 mL of .875 M HNO; is combined with 100.0 mL of 0.167 M NaOH what is the final pH? (Assume that the volumes are additive.) (d) 1.38 b) 2.28 c) 11.72 d) 12.62 strong

Answer 1

Given:

M(HNO3) = 0.875 M

V(HNO3) = 25 mL

M(NaOH) = 0.167 M

V(NaOH) = 100 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.875 M * 25 mL = 21.875 mmol

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

mol(NaOH) = 0.167 M * 100 mL = 16.7 mmol

We have:

mol(HNO3) = 21.88 mmol

mol(NaOH) = 16.7 mmol

16.7 mmol of both will react

remaining mol of HNO3 = 5.175 mmol

Total volume = 125.0 mL

[H+]= mol of acid remaining / volume

[H+] = 5.175 mmol/125.0 mL

= 4.14*10^-2 M

use:

pH = -log [H+]

= -log (4.14*10^-2)

= 1.38

Answer: 1.38