Question

29.91 ml of an acid solution (identity unknown) is neutralized by 17.29 ml of 0.1999 m calcium hydroxide solution. determine the normality of the acid.

Explanation:

Step1: Find normality of $\text{Ca(OH)}_2$

Calcium hydroxide provides 2 $\text{OH}^-$ ions, so normality $N = 2 \times M$.
$N_{\text{base}} = 2 \times 0.1999 = 0.3998\ \text{N}$

Step2: Use neutralization equivalence law

At neutralization, $N_{\text{acid}}V_{\text{acid}} = N_{\text{base}}V_{\text{base}}$. Rearrange to solve for $N_{\text{acid}}$:
$N_{\text{acid}} = \frac{N_{\text{base}}V_{\text{base}}}{V_{\text{acid}}}$

Step3: Substitute values into formula

$N_{\text{acid}} = \frac{0.3998 \times 17.29}{29.91}$
Calculate numerator: $0.3998 \times 17.29 \approx 6.9125$
Then divide: $\frac{6.9125}{29.91} \approx 0.2311$

Answer:

$0.2311\ \text{N}$