Question

titration example: calculate the volume (in ml) of a 1.5 m naoh solution required to titrate 50.0 ml of a 0.045 m h₂co₃ solution (i.e., 1 can of soda)?

Explanation:

Step1: Write the balanced chemical equation

$$\ce{H_{2}CO_{3} + 2NaOH ightarrow Na_{2}CO_{3} + 2H_{2}O}$$

From the equation, the mole ratio of $\ce{H_{2}CO_{3}}$ to $\ce{NaOH}$ is $n_{\ce{H_{2}CO_{3}}}:n_{\ce{NaOH}} = 1:2$.

Step2: Calculate the number of moles of $\ce{H_{2}CO_{3}}$

Use the formula $n = C\times V$ (where $C$ is concentration and $V$ is volume).
Given $C_{\ce{H_{2}CO_{3}}}=0.045\ \text{M}$ and $V_{\ce{H_{2}CO_{3}}}=50.0\ \text{mL}=0.050\ \text{L}$
$$n_{\ce{H_{2}CO_{3}}}=C_{\ce{H_{2}CO_{3}}}\times V_{\ce{H_{2}CO_{3}}}=0.045\ \text{mol/L}\times0.050\ \text{L}=0.00225\ \text{mol}$$

Step3: Calculate the number of moles of $\ce{NaOH}$

Since $n_{\ce{H_{2}CO_{3}}}:n_{\ce{NaOH}} = 1:2$, then $n_{\ce{NaOH}} = 2\times n_{\ce{H_{2}CO_{3}}}$
$$n_{\ce{NaOH}}=2\times0.00225\ \text{mol}=0.0045\ \text{mol}$$

Step4: Calculate the volume of $\ce{NaOH}$ solution

Use the formula $V=\frac{n}{C}$ (re - arranged from $n = C\times V$). Given $C_{\ce{NaOH}} = 1.5\ \text{M}$ and $n_{\ce{NaOH}}=0.0045\ \text{mol}$
$$V_{\ce{NaOH}}=\frac{n_{\ce{NaOH}}}{C_{\ce{NaOH}}}=\frac{0.0045\ \text{mol}}{1.5\ \text{mol/L}}=0.003\ \text{L}$$
Convert liters to milliliters: $V_{\ce{NaOH}}=0.003\ \text{L}\times1000\ \text{mL/L}=3\ \text{mL}$

Answer:

$3\ \text{mL}$