Chapter 10: Q. 10.95 (page 360)

A 0,205 M NaOH solution is used to titrate $20.0 mL$ of a solution of $H_{2} {SO}_{4}$ . If $45.6 ml$ , of the $NaOH$ solution is required to reach the endpoint, what is the molarity of the $H_{2} {SO}_{4}$ solution? $(0, 7)$
$H_{2} {SO}_{4} (a q) + 2 NaOH (α q) ⟶ 2 H_{2} OND) + {Na}_{2} {SO}_{4} (α q)$

Short Answer

Expert verified

Molarity $H_{2} {SO}_{4} = \frac{2 \times 0.00234 {moleH}_{2} {SO}_{4}}{0.02 L}$

$= 0.234 {MH}_{2} {SO}_{4}$

Step by step solution

Given information

That are given numbers are calculated by : (1)

Metric Molarity Mole divide by

Explanation

Step 1 Given $45.6 ml$ of $0.205 MNaOH; 20.0 {mlH}_{2} {SO}_{4} = 0.02 {LH}_{2} {SO}_{4}$ .

Need molarity of $H_{2} {SO}_{4}$

Step 2 Plan

Metric Molarity Mole divide by

$45.6 mlL$ moles $NaOH$ moles $H_{2} {SO}_{4} {MH}_{2} {SO}_{4}$

Factor factor factor liters

Step 3 Equalities /conversion factors

1 L NaOH = 0.205 moleNaOH \frac{1 L}{0.205 mole} = \frac{0.205 mole}{1 L} 1 L NaOH = 1000 mlNaOH \frac{1 L}{1000 ml} = \frac{1000 ml}{1 L}

Given information

Equation are:

Explanation

2 mole $NaOH =$ l mole $H_{2} {SO}_{4}$

$\frac{2 m o l e N a O H}{1 m o l e H_{2} {SO}_{4}}$ and $\frac{Imole H_{2} {SO}_{4}}{2 mole NaOH}$

Step 4 Set up problem

$45.6 mlSeOH \times \frac{1 LNeOH}{1000 mlNaOH} \times \frac{0.205 moleNeOH}{1 L SaeOH} \times \frac{lmole H_{2} {SO}_{4}}{2 mole, SeOH} = 0.00234 {moleH}_{2} {SO}_{4} Molarity H_{2} {SO}_{4} = \frac{2 \times 0.00234 {mole}_{2} {SO}_{4}}{0.02 L} = 0.234 {MH}_{2} {SO}_{4}$

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A 0,205 M NaOH solution is used to titrate 20.0mLof a solution of H2SO4. If 45.6ml, of the NaOHsolution is required to reach the endpoint, what is the molarity of the H2SO4solution? (0,7)H2SO4(aq)+2NaOH(αq)⟶2H2OND+Na2SO4(αq)

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