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Chapter 6: Problem 135

What volume of \(0.0521 M B a(O H)_{2}\) is required to neutralize exactly \(14.20 \mathrm{mL}\) of \(0.141 M \mathrm{H}_{3} \mathrm{PO}_{4} ?\) Phosphoric acid contains three acidic hydrogens.

Short Answer

Expert verified
The volume of $0.0521 \, M \, Ba(OH)_2$ required to neutralize exactly $14.20 \, mL$ of $0.141 \, M \, H_{3}PO_{4}$ is $25.52 \, mL$.

Step by step solution

01

Write the balanced neutralization reaction

Write the balanced chemical equation for the reaction between barium hydroxide (Ba(OH)2) and phosphoric acid (H3PO4). \[ 2 Ba(OH)_{2} + 3 H_{3}PO_{4} \rightarrow Ba_{3}(PO_{4})_{2} + 6 H_{2}O \]
02

Identify known quantities

We're given the following information: - Molarity of Ba(OH)2 = 0.0521 M - Volume of H3PO4 = 14.20 mL - Molarity of H3PO4 = 0.141 M
03

Calculate moles of H3PO4

Use the molarity and volume of H3PO4 to calculate its moles in the given solution. Molarity (moles/Liter) = moles/volume(L). \[ moles\_of\_H_{3}PO_{4} = Molarity\_of\_H_{3}PO_{4} \times \frac{Volume\_of\_H_{3}PO_{4}}{1000} \] \[ moles\_of\_H_{3}PO_{4} = 0.141 \times \frac{14.20}{1000} \] \[ moles\_of\_H_{3}PO_{4} = 2.00 \times 10^{-3} \, moles \]
04

Calculate moles of Ba(OH)2 using stoichiometry

Use stoichiometry to find the moles of Ba(OH)2 needed to neutralize H3PO4 completely. From the balanced equation, 2 moles of Ba(OH)2 reacts with 3 moles of H3PO4. \[ \frac{moles\_of\_Ba(OH)_{2}}{moles\_of\_H_{3}PO_{4}} = \frac{2}{3} \] Solve for moles of Ba(OH)2: \[ moles\_of\_Ba(OH)_{2} = \frac{2}{3} \times moles\_of\_H_{3}PO_{4} \] \[ moles\_of\_Ba(OH)_{2} = \frac{2}{3} \times 2.00 \times 10^{-3} \] \[ moles\_of\_Ba(OH)_{2} = 1.33 \times 10^{-3} \, moles \]
05

Calculate volume of Ba(OH)2 required

Now with the moles of Ba(OH)2 and its molarity, we can find the required volume of the Ba(OH)2 solution. Rearrange the molarity equation to find the volume. \[ volume\_of\_Ba(OH)_{2} = \frac{moles\_of\_Ba(OH)_{2}}{Molarity\_of\_Ba(OH)_{2}} \times 1000 \] Plug in the values: \[ volume\_of\_Ba(OH)_{2} = \frac{1.33 \times 10^{-3}}{0.0521} \times 1000 \] \[ volume\_of\_Ba(OH)_{2} = 25.52 \, mL \] Therefore, 25.52 mL of 0.0521 M Ba(OH)2 solution is required to neutralize exactly 14.20 mL of 0.141 M H3PO4.

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