Question

Determine the pH of a solution when 23.2 mL of 0.026 M HNO3 is mixed with 15.5 mL of A) 0.0140 M NaOH.

A) 0.0140 M NaOH

B)distilled water

C)0.0070 M HCl

D)0.186 M KOH

Answer 1

HCl + NaOH =⇒ NaCl + H2O

0.026mole HNO3 x 23.2 mL x 1 liter /1000 mL = 6.032e-4 moles
A)

0.0140 moles NaOH x 15.5 mL / 1000 mL*1liter = 2.17e-4 moles
moles of HNO3 that did not react:
6.032e-4 – 2.17e-4 =0.000386 moles
Molarity of HNO3 = 0.000386 moles/23.2 + 15.5 mL / 1000 mL/1liter = 0.0155 M
HNO3 =⇒ H+ + NO3-
pH = - log[H+]
pH = - log(0.0155) = -(-1.81) = 1.81

B) 6.032e-4 moles HNO3/ 23.2 + 15.5 mL / 1000 mL= 0.015526
pH = -log(0.015526) = -(-1.808) = 1.808

C) 0.0070 moles HCl/1liter x 23.2 mL x 1liter/1000 mL = 0.000162 moles HCl
HCl =⇒ H+ + Cl-
Total moles H+ = 0.000162 moles of from HNO3 + 0.000162 from HCl = .00076
[H+] = 0.00076 moles / 23.2 + 15.5 mL / 1000 mL = 0.0155
pH = -log(0.0155) = -(-1.808) = 1.808

D) you can do it as usual