Question

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

Explanation:

Step1: Write the balanced chemical equation

$$\ce{H2SO4 + Ba(OH)2 -> BaSO4 + 2H2O}$$
From the equation, the mole ratio of $\ce{H2SO4}$ 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).
Given $M = 1.885\ M$ and $V=12.15\ mL=12.15\times10^{- 3}\ L$
$$n_{\ce{Ba(OH)2}}=1.885\ M\times12.15\times10^{-3}\ L = 0.02290275\ mol$$

Step3: Determine the moles of $\ce{H2SO4}$

Since the mole ratio of $\ce{H2SO4}$ to $\ce{Ba(OH)2}$ is $1:1$, $n_{\ce{H2SO4}}=n_{\ce{Ba(OH)2}} = 0.02290275\ mol$

Step4: Calculate the molarity of $\ce{H2SO4}$

Given $V_{\ce{H2SO4}} = 25\ mL=25\times10^{-3}\ L$
Use the formula $M=\frac{n}{V}$
$$M_{\ce{H2SO4}}=\frac{0.02290275\ mol}{25\times 10^{-3}\ L}=0.91611\ M\approx0.916\ M$$

Answer:

The concentration of $\ce{H2SO4}$ is approximately $0.916\ M$