Question

② 25 ml h₂so₄ reacting with 12.15 ml of 1.885 m ba(oh)₂
find conc. h₂so₄
formula = moles = molarity × volume
h₂so₄ = 25 ml

Explanation:

Step1: Write the balanced chemical equation

$$\ce{H_{2}SO_{4} + Ba(OH)_{2}=BaSO_{4}\downarrow + 2H_{2}O}$$
From the equation, the mole ratio of $\ce{H_{2}SO_{4}}$ to $\ce{Ba(OH)_{2}}$ is $1:1$.

Step2: Calculate the moles of $\ce{Ba(OH)_{2}}$

Use the formula $n = M\times V$ (where $n$ is moles, $M$ is molarity, and $V$ is volume in liters).
$V_{Ba(OH)_{2}}=12.15\space mL = 12.15\times10^{- 3}\space L$, $M_{Ba(OH)_{2}} = 1.885\space M$
$n_{Ba(OH)_{2}}=M\times V=1.885\space M\times12.15\times 10^{-3}\space L$
$n_{Ba(OH)_{2}}=1.885\times12.15\times10^{-3}\space mol\approx0.0229\space mol$

Step3: Determine the moles of $\ce{H_{2}SO_{4}}$

Since the mole ratio of $\ce{H_{2}SO_{4}}$ to $\ce{Ba(OH)_{2}}$ is $1:1$, $n_{H_{2}SO_{4}}=n_{Ba(OH)_{2}}\approx0.0229\space mol$

Step4: Calculate the molarity of $\ce{H_{2}SO_{4}}$

$V_{H_{2}SO_{4}} = 25\space mL=25\times10^{-3}\space L$
Use the formula $M=\frac{n}{V}$
$M_{H_{2}SO_{4}}=\frac{n_{H_{2}SO_{4}}}{V_{H_{2}SO_{4}}}=\frac{0.0229\space mol}{25\times10^{-3}\space L}$
$M_{H_{2}SO_{4}} = 0.916\space M$

Answer:

The concentration of $\ce{H_{2}SO_{4}}$ is $0.916\space M$