Question

what volume of a 0.107 m nitric acid solution is required to neutralize 11.3 ml of a 0.134 m barium hydroxide solution?
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1 ml nitric acid
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Explanation:

Step1: Write the balanced chemical equation

$$2HNO_3 + Ba(OH)_2=Ba(NO_3)_2 + 2H_2O$$
From the equation, the mole ratio of \(HNO_3\) to \(Ba(OH)_2\) is \(n_{HNO_3}:n_{Ba(OH)_2}=2:1\)

Step2: Calculate the moles of \(Ba(OH)_2\)

Use the formula \(n = C\times V\) (where \(C\) is concentration and \(V\) is volume in liters).
\(V_{Ba(OH)_2}=11.3\space mL=11.3\times10^{- 3}\space L\), \(C_{Ba(OH)_2}=0.134\space M\)
\(n_{Ba(OH)_2}=C_{Ba(OH)_2}\times V_{Ba(OH)_2}=0.134\space M\times11.3\times 10^{-3}\space L = 1.5142\times10^{-3}\space mol\)

Step3: Calculate the moles of \(HNO_3\)

Since \(n_{HNO_3}=2n_{Ba(OH)_2}\)
\(n_{HNO_3}=2\times1.5142\times 10^{-3}\space mol=3.0284\times10^{-3}\space mol\)

Step4: Calculate the volume of \(HNO_3\)

Use the formula \(V=\frac{n}{C}\) (where \(n\) is moles and \(C\) is concentration).
\(C_{HNO_3}=0.107\space M\), \(n_{HNO_3}=3.0284\times10^{-3}\space mol\)
\(V_{HNO_3}=\frac{n_{HNO_3}}{C_{HNO_3}}=\frac{3.0284\times 10^{-3}\space mol}{0.107\space mol/L}=28.3\space mL\)

Answer:

\(28.3\space mL\)