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Chapter 4: Problem 125

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

Short Answer

Expert verified
To neutralize exactly \(14.20 \ \text{mL}\) of \(0.141 \ \text{M} \ \text{H}_{3}\text{PO}_{4}\), approximately \(0.0256 \ \text{L}\) (or \(25.6 \ \text{mL}\)) of \(0.0521 \ \text{M} \ \text{Ba(OH)}_{2}\) solution is required.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the reaction between barium hydroxide and phosphoric acid is: \(2 \mathrm{Ba(OH)_2 + 3H_3PO_4 \rightarrow Ba_3(PO_4)_2 + 6H_2O\)
02

Determine moles of H3PO4

First, we need to determine the moles of H3PO4 in the given volume. We can do this using the given concentration and volume. The formula is: Moles of H3PO4 = (concentration of H3PO4) 脳 (volume of H3PO4) Moles of H3PO4 = (0.141 mol/L) 脳 (14.20 mL 脳 (1 L / 1000 mL)) Moles of H3PO4 = 0.0020022 mol
03

Use stoichiometry to determine moles of Ba(OH)2

From the balanced chemical equation, we know that 2 moles of Ba(OH)2 are needed to neutralize 3 moles of H3PO4. Using this stoichiometric ratio, we can determine the moles of Ba(OH)2 required: You have \(\frac{2}{3}\) moles of \(Ba(OH)_2\) per mole of \(H_3PO_4\) Moles of Ba(OH)2 = Moles of H3PO4 脳 \(\frac{2 \ \text{moles Ba(OH)2}}{3 \ \text{moles H3PO4}}\) Moles of Ba(OH)2 = 0.0020022 mol 脳 \(\frac{2}{3}\) = 0.0013348 mol
04

Calculate the volume of Ba(OH)2

Now we can use the concentration of Ba(OH)2 to determine the volume required. The formula is: Volume of Ba(OH)2 = \(\frac{\text{Moles of Ba(OH)2}}{\text{Concentration of Ba(OH)2}}\) Volume of Ba(OH)2 = \(\frac{0.0013348 \ \text{mol}}{0.0521 \ \text{mol/L}}\) Volume of Ba(OH)2 鈮� 0.0256 L

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Most popular questions from this chapter

Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. It is composed solely of \(\mathrm{C}, \mathrm{F}\), and \(\mathrm{B}\); it is \(42.23 \% \mathrm{C}\) by mass and \(55.66 \% \mathrm{~F}\) by mass. a. What is the empirical formula of BARF? b. A \(2.251-g\) sample of BARF dissolved in \(347.0 \mathrm{~mL}\) of solution produces a \(0.01267 M\) solution. What is the molecular formula of BARF?

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