0 cm³ of 0.2 M solution of hydrochloric acid required 12.5 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of sodium hydroxide.

The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is a neutralization reaction.

The mole ratio of HCl to NaOH is $1:1$.

Given:
Volume of HCl ($V_{\text{HCl}}$) = 25.0 cm³ = $25.0 \times 10^{-3}$ L = 0.025 L
Concentration of HCl ($M_{\text{HCl}}$) = 0.2 M (mol/L)

Moles of HCl ($n_{\text{HCl}}$) = $M_{\text{HCl}} \times V_{\text{HCl}}$

From the balanced equation, 1 mole of HCl reacts with 1 mole of NaOH.
So, moles of NaOH ($n_{\text{NaOH}}$) = moles of HCl = 0.005 mol.

Given:
Volume of NaOH ($V_{\text{NaOH}}$) = 12.5 cm³ = $12.5 \times 10^{-3}$ L = 0.0125 L

Concentration of NaOH ($M_{\text{NaOH}}$) = $\frac{n_{\text{NaOH}}}{V_{\text{NaOH}}}$

(i) The concentration of sodium hydroxide in moles per litre is:

Molar mass of NaOH:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Molar mass of NaOH = $22.99 + 16.00 + 1.01 = 40.00 \text{ g/mol}$

Concentration (g/L) = Concentration (mol/L) $\times$ Molar mass (g/mol)
Concentration of NaOH (g/L) = $0.4 \text{ mol/L} \times 40.00 \text{ g/mol} = 16.0 \text{ g/L}$

(ii) The concentration of sodium hydroxide in grams per litre is:

Question 2:
About 100 cm³ of a solution of hydrochloric acid reacted completely with 0.12g of magnesium ribbon. Calculate the concentration of the hydrochloric acid in:

(i) moles per litre.

(ii) grams per litre.

The reaction between magnesium (Mg) and hydrochloric acid (HCl) produces magnesium chloride and hydrogen gas.

The mole ratio of Mg to HCl is $1:2$.

Given:
Mass of Mg = 0.12 g
Molar mass of Mg = 24.31 g/mol

Moles of Mg ($n_{\text{Mg}}$) = $\frac{\text{mass of Mg}}{\text{molar mass of Mg}}$

From the balanced equation, 1 mole of Mg reacts with 2 moles of HCl.
Moles of HCl ($n_{\text{HCl}}$) = $2 \times n_{\text{Mg}}$

Given:
Volume of HCl ($V_{\text{HCl}}$) = 100 cm³ = $100 \times 10^{-3}$ L = 0.100 L

Concentration of HCl ($M_{\text{HCl}}$) = $\frac{n_{\text{HCl}}}{V_{\text{HCl}}}$

Rounding to two significant figures (based on 0.12g):

(i) The concentration of hydrochloric acid in moles per litre is:

Molar mass of HCl:
H = 1.01 g/mol
Cl = 35.45 g/mol
Molar mass of HCl = $1.01 + 35.45 = 36.46 \text{ g/mol}$

Concentration (g/L) = Concentration (mol/L) $\times$ Molar mass (g/mol)
Concentration of HCl (g/L) = $0.09872 \text{ mol/L} \times 36.46 \text{ g/mol} \approx 3.606 \text{ g/L}$
Rounding to two significant figures:

(ii) The concentration of hydrochloric acid in grams per litre is: