Question

to determine the concentration of a solution of nitrous acid, a 100.0 - ml sample is placed in a flask and titrated with a 0.1132 m solution of sodium hydroxide. a volume of 33.68 ml is required to reach the phenolphthalein endpoint. calculate the concentration of nitrous acid in the original sample. 3 item attempts remaining

Explanation:

Step1: Write the balanced chemical equation

The reaction between nitrous acid ($HNO_2$) and sodium hydroxide ($NaOH$) is $HNO_2+NaOH = NaNO_2 + H_2O$. The mole - ratio of $HNO_2$ to $NaOH$ is 1:1.

Step2: Calculate the moles of $NaOH$ used

Use the formula $n = M\times V$, where $M$ is the molarity and $V$ is the volume in liters. The volume of $NaOH$ is $V_{NaOH}=33.68\ mL = 0.03368\ L$ and its molarity $M_{NaOH}=0.1132\ M$. So, $n_{NaOH}=M_{NaOH}\times V_{NaOH}=0.1132\ mol/L\times0.03368\ L = 0.003802\ mol$.

Step3: Determine the moles of $HNO_2$

Since the mole - ratio of $HNO_2$ to $NaOH$ is 1:1, $n_{HNO_2}=n_{NaOH}=0.003802\ mol$.

Step4: Calculate the molarity of $HNO_2$

The volume of the $HNO_2$ sample is $V_{HNO_2}=100.0\ mL = 0.1000\ L$. Using the formula $M=\frac{n}{V}$, we have $M_{HNO_2}=\frac{n_{HNO_2}}{V_{HNO_2}}=\frac{0.003802\ mol}{0.1000\ L}=0.03802\ M$.

Answer:

$0.03802\ M$