Question

an aqueous solution of nitric acid is standardized by titration with a 0.134 m solution of barium hydroxide.
if 11.3 ml of base are required to neutralize 10.5 ml of the acid, what is the molarity of the nitric acid solution?
0.144 m nitric acid
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Explanation:

Step1: Write the balanced chemical equation

$$ m Ba(OH)_2 + 2HNO_3=Ba(NO_3)_2 + 2H_2O$$

The mole ratio of $
m Ba(OH)_2$ to $
m HNO_3$ is $1:2$.

Step2: Calculate the moles of $

m Ba(OH)_2$
Use the formula $n = M\times V$ (where $n$ is moles, $M$ is molarity, and $V$ is volume in liters).
$V_{Ba(OH)_2}=11.3\space mL = 11.3\times10^{- 3}\space L$, $M_{Ba(OH)_2}=0.134\space M$
$n_{Ba(OH)_2}=M\times V=0.134\space M\times11.3\times 10^{-3}\space L = 0.0015142\space mol$

Step3: Calculate the moles of $

m HNO_3$
From the mole ratio ($n_{HNO_3} = 2n_{Ba(OH)_2}$)
$n_{HNO_3}=2\times0.0015142\space mol=0.0030284\space mol$

Step4: Calculate the molarity of $

m HNO_3$
$V_{HNO_3}=10.5\space mL=10.5\times 10^{-3}\space L$
Use the formula $M=\frac{n}{V}$
$M_{HNO_3}=\frac{0.0030284\space mol}{10.5\times 10^{-3}\space L}=0.2884\space M\approx0.288\space M$

Answer:

$0.288\space M$