Question

1. What volume of a 0.200 M calcium hydroxide solution is required to neutralize 19.8 mL of a 0.144 M perchloric acid solution?

__mL calcium hydroxide

2. What volume of a 0.102 M nitric acid solution is required to neutralize 13.0 mL of a 0.143 M sodium hydroxide solution?

___mL nitric acid

Answer 1

Question 1.

The chemical reaction between calcium hydroxide and perchloric acid can be written as

$$Ca{(OH)_2} + 2HCl{O_4} \to Ca{(Cl{O_4})_2} + 2{H_2}O$$

Thus we see, that they react 1 to 2 (1 mole of calcium hydroxide to 2 moles of perchloric acid). The number of moles of calcium hydroxide we need to neutralize can be calculated as ${\nu _{HCl{O_4}}} = {C_{HCl{O_4}}} \cdot {V_{HCl{O_4}}}$ , the number of moles of calcium hydroxide is the half of ${\nu _{HCl{O_4}}}$ , thus we can write

$${C_{Ca{{(OH)}_2}}} \cdot {V_{Ca{{(OH)}_2}}} = {\nu _{Ca{{(OH)}_2}}} = {{{\nu _{HCl{O_4}}}} \over 2} = {1 \over 2}{C_{HCl{O_4}}} \cdot {V_{HCl{O_4}}}$$

Then

$${V_{Ca{{(OH)}_2}}} = {1 \over 2}{{{C_{HCl{O_4}}}} \over {{C_{Ca{{(OH)}_2}}}}} \cdot {V_{HCl{O_4}}}$$

Let's do the calculations

$${V_{Ca{{(OH)}_2}}} = {1 \over 2}{{0.144[{\rm{M}}]} \over {0.200[{\rm{M}}]}} \cdot 19.8[{\rm{mL}}] = 7.128[{\rm{mL}}]$$

Answer: ${V_{Ca{{(OH)}_2}}} = 7.128[{\rm{mL}}]$

Question 2.

The chemical reaction between nitric acid and sodium hydroxide can be written as

$$NaOH + HN{O_3} = NaN{O_3} + {H_2}O$$

All reasoning are the same as in Question 1 except now they react 1 to 1, thus

$${V_{HN{O_3}}} = {{{C_{NaOH}}} \over {{C_{HN{O_3}}}}} \cdot {V_{NaOH}}$$ $${V_{HN{O_3}}} = {{0.143[{\rm{M}}]} \over {0.102[{\rm{M}}]}} \cdot 13.0[{\rm{mL}}] \approx 18.2255[{\rm{mL}}]$$