0.02268L x 0.015M Sr(OH)2 = 0.0034mol Sr(OH)2 Sr(OH)2 +

ACTIVITY #3 TITRATION PRACTICE QUESTIONS
1. A student used a 22.68 mL of 0.015 M Sr(OH)2 to titrate 5.00 mL of HNO3. Calculate the
concentration of HNO3.
0.02268L x 0.015M Sr(OH)2 = 0.0034mol Sr(OH)2
Sr(OH)2 + 2HNO3 β†’ Sr(NO3)2 + 2H2O
2 𝐻𝑁𝑂3
0.0034 mol Sr(OH)2 x
1 π‘†π‘Ÿ(𝑂𝐻)2
= 0.00068mol
0.00068 mol / 0.005L = 0.13M HNO3
2. What volume of 0.50M H2C2O4 is required to titrate 10.00mL of 0.12M NaOH?
0.010L x 0.12M NaOH = 0.0012mol NaOH
2NaOH + H2C2O4 β†’ Na2C2O4 + 2H2O
0.0012 mol NaOH x
0.0006 mol x
1 𝐻2𝐢2𝑂4
2 π‘π‘Žπ‘‚π»
1𝐿
0.50 π‘šπ‘œπ‘™
= 0.0006 mol
= 0.0012L = 1.2mL H2C2O4
3. A student titrated 25.00mL of HCl with 15.62mL of 0.30M NaOH. Calculate the
concentration of HCl.
0.01562L x 0.30M NaOH = 0.0047mol NaOH
NaOH + HCl→ NaCl + H2O
0.0047 mol NaOH x
1 𝐻𝐢𝑙
1 π‘π‘Žπ‘‚π»
= 0.0047 mol
0.0047 mol / 0.025L = 0.188M HCl
4. In a titration, a 10.00mL sample of NaOH was titrated with 24.25mL of 0.20M H2SO4.
Calculate the concentration of NaOH.
0.02425 L x 0.20M H2SO4 = 0.00485 mol H2SO4
2NaOH + H2SO4 β†’Na2SO4 + 2H2O
0.00485 mol x
2 π‘π‘Žπ‘‚π»
1 𝐻2𝑆𝑂4
= 0.0097 mol NaOH
0.0097mol / 0.0100L = 0.97M NaOH