Question

consider the rotation of 50.0mL of a 0.2M HNO3 with 0.10M NaOH calculate the pH of the solution

a) before any extra NaOH has been add
b) after 50mL of NaOH
c) add 100mL of NaOH
d) add 150mL of NaOH

Answer 1

a)

before addition means there is only HNO3 and it is strong acid

pH = -log(0.2) = 0.7

b)

after 50 mL 0.1 M NaOH

moles of HNO3 = Molarity x volume of the solution in Liters

= 0.2 M x 0.05 L = 0.01 mol

Molaes of NaOH = 0.1M x 0.05 L = 0.005 mol

0.005 mol NaOH react with 0.005 mol HNO3

moles of HNO3 remaining = 0.01 mol - 0.005 mol = 0.005 mol

total volume = 50+50 = 100 mL = 0.1L

concentration of HNO3 remaining = 0.005 mol / 0.1 L = 0.05 M

pH = -log0.05 = 1.3

c)

100 mL of NaOH

moles of NaOH = 0.1M x 0.1L = 0.01 mol NaOH

0.01 molNaOH react with 0.01 mol HNO3

moles of HNO3 remaining = 0.01 - 0.01 = 0

means exactly at nuetral point so pH = 7

d)

150 mL of NaOH

moles = 0.1 M x 0.15L = 0.015 mol

0.01 ol of HNO3 react with 0.01 moles of NaOH

moles of NaOH remaining = 0.015 - 0.01 = 0.005 mol

total volume = 150 + 50 = 200mL = 0.2L

concentration of remaining NaOH = 0.005 mol / 0.2 L = 0.025 M

since NaOH is strong base

pOH = -log0.025 = 1.6

pH = 14-pOH = 14 - 1.6 = 12.4