Question

25cm³ of a solution containing 4.4grams/litre sodium hydroxide was required to neutralize 17.8cm²of a propanoic acid . Determine the concentration of the propanoic acid in gram's/litre?

Answer 1

$C_2H_5OOH + NaOH \to C_2H_5OONa + H_2O$

Since the molar mass of NaOH = 74g/mol

Molar concentration of NaOH = $\dfrac{\textsf{mass concentration (g/L)}}{\textsf{molar mass (g/mol)}} = \dfrac{4.4g/L}{40g/mol}$ = 0.11 mol/L

Using the formula,

$\dfrac{c_av_a}{c_bv_b}= \dfrac{n_a}{n_b}$

$\dfrac{c_a × 17.8}{0.110×25} = \dfrac{1}{1}$

c_a = $\dfrac{0.110×25}{17.8}$ = 0.154mol/L

since the molar mass of propanoic acid = 74g/mol

mass concentration of propanoic acid = molar concentration (mol/L) × molar mass (g/mol) = = 0.154g/L × 74g/mol = 11.4g/L

Therefore, the concentration of the propanoic acid in grammes/litre is 11.4g/L.