Question

38.0 mL of 1.20 M perchloric acid is added to 16.9 mL of sodium hydroxide, and the resulting solution is found to be acidic. 15.9 mL of 0.950 M calcium hydroxide is required to reach neutrality. What is the molarity of the original sodium hydroxide solution? _____ M

Answer 1

0.9107M

Explanation

The reaction between HClO4 and sodium hydroxide is 1:1molar

HClO4(aq) + NaOH(aq) -------> NaClO4(aq) + H2O(l)

The reaction between HClO4 and Ca(OH)2 is 2:1 molar

2HClO4(aq) + Ca(OH)2(aq) ------> Ca(ClO4)2 (aq) + 2H2O(l)

moles of HClO4 added = (1.20mol/1000ml) × 38.0ml = 0.0456mol

moles of Ca(OH)2 added = ( 0.950mol/1000ml ) ×15.9ml = 0.015105mol

moles of HClO4 reacted with Ca(OH)2 = 0.015105mol × 2 = 0.03021mol

moles of HClO4 consumed for NaOH = 0.0456mol - 0.03021mol = 0.01539mol

moles of NaOH present in the 16.9 ml solution = (0.01539mol/16.9ml) ×1000ml = 0.9107M