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50 ml solution of 0.1 m H2SO was titrated with 0.2 m NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.5 m KOH solution. The volume (in ml) of KOH required for completing the titration is
Accepted Answer
Assuming that the neutralization reaction between H2SO4 and NaOH goes to completion, the number of moles of H2SO4 initially present in 50 mL of 0.1 M solution is:
0.1 mol/L * 0.050 L = 0.005 mol H2SO4
Since H2SO4 is a diprotic acid, it will react with 2 moles of NaOH. Therefore, the number of moles of NaOH required for complete neutralization is:
0.005 mol H2SO4 * 2 mol NaOH/1 mol H2SO4 = 0.010 mol NaOH
Since 30 mL of 0.2 M NaOH solution has already been added, the number of moles of NaOH added so far is:
0.2 mol/L * 0.030 L = 0.006 mol NaOH
Therefore, the number of moles of NaOH still required to complete the neutralization is:
0.010 mol NaOH - 0.006 mol NaOH = 0.004 mol NaOH
Finally, the volume of 0.5 M KOH solution required to provide 0.004 mol of NaOH is:
0.004 mol NaOH * (1 L/0.5 mol NaOH) = 0.008 L = 8 mL
Therefore, the volume of KOH required for completing the titration is 8 mL.
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