Question 7.2.2: Additional Question Options

2022 - May/June - Paper 2 - Question 7

Or

Option 2

\frac{c_1V_1}{c_2V_2}=\frac{n_1}{n_2}

\frac{\mathrm{c}_1(200)}{(0,01)(350)}=\frac11

\mathrm{c}_1=0,0175\mathrm{~mol}\cdot\mathrm{dm}^{-3}

\mathrm{c}(\mathrm{HCl})_{\text{react }}=\mathrm{c}(\mathrm{HCl})_{\text{ini }}-\mathrm{c}(\mathrm{HCl})_{\text{excess }}

=0,03-0,0175

=0.0125\mathrm{~mol}\cdot\mathrm{dm}^{-3}

\frac{\mathrm{c}_{\mathrm{a}}V_{\mathrm{a}}}{\mathrm{c}_{\mathrm{b}}V_{\mathrm{b}}}=\frac{\mathrm{n}_{\mathrm{a}}}{\mathrm{n}_{\mathrm{b}}}

\frac{(0,0125)(200)}{c_{b}(150)}=\frac11

\mathrm{c}(\mathrm{NaOH})=0,0167\mathrm{~mol}\cdot\mathrm{dm}^{-3}

\begin{aligned} & \left(0,0167\mathrm{mol}\cdot\mathrm{dm}^{-3}\right.\\ & \text{ or }0,017\mathrm{mol}\cdot\mathrm{dm}^{-3}\text{ \rparen}\end{aligned}

Or

Option 3

Concentration ratio in final solution:

\mathrm{HCl}:\mathrm{H}_3\mathrm{O}^{+}=1:1

Thus:

[\mathrm{HCl}]=0,01\mathrm{~mol}\cdot\mathrm{dm}^{-3}

[\mathrm{HCl}]_{\text{react }}=[\mathrm{HCl}]_{\text{initial }}-[\mathrm{HCl}]_{\text{excess }}

=0,03-0,01

=0,02\mathrm{~mol}\cdot\mathrm{dm}^{-3}

Concentration ratio in final solution:

\mathrm{HCl}:\mathrm{NaOH}=1:1

[\mathrm{NaOH}]=0,02\mathrm{~mol}\cdot\mathrm{dm}^{-3}

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