Question

question 5

if you added 2.00 grams of solid naoh to a 250.0 ml solution of 0.100 molar hydrochloric acid, what would be the ph of the resulting solution after all the naoh had dissolved?

hint: 1 mole of naoh will neutralize (cancel out) 1 mole of hcl.

assume the naoh added has no effect on the total volume of the solution.

Explanation:

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<reasoning>

Calculate moles of reactants

Using the Molar Mass Calculation and Molarity knowledge points
\[

$$\begin{aligned} \text{Molar mass of NaOH} &= 22.99 + 16.00 + 1.01 = 40.00\text{ g/mol}\\ n_{\text{NaOH}} &= \frac{2.00\text{ g}}{40.00\text{ g/mol}} = 0.0500\text{ mol}\\ n_{\text{HCl}} &= 0.2500\text{ L} \times 0.100\text{ M} = 0.0250\text{ mol} \end{aligned}$$

\]

Determine excess reactant after neutralization

Using the Acid-Base Neutralization Stoichiometry knowledge point
\[

$$\begin{aligned} \text{Reaction:} \quad &\text{NaOH} + \text{HCl} ightarrow \text{NaCl} + \text{H}_2\text{O}\\ n_{\text{NaOH, excess}} &= 0.0500\text{ mol} - 0.0250\text{ mol} = 0.0250\text{ mol} \end{aligned}$$

\]

Calculate hydroxide ion concentration

Using the Molarity knowledge point
\[

$$\begin{aligned} [\text{OH}^-] &= \frac{0.0250\text{ mol}}{0.2500\text{ L}} = 0.100\text{ M} \end{aligned}$$

\]

Calculate pH of the solution

Using the pH and pOH Calculations knowledge point
\[

$$\begin{aligned} \text{pOH} &= -\log_{10}(0.100) = 1.00\\ \text{pH} &= 14.00 - 1.00 = 13.00 \end{aligned}$$

\]
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<answer>
If you added 2.00 grams of solid NaOH to a 250.0 mL solution of 0.100 molar hydrochloric acid, what would be the pH of the resulting solution after all the NaOH had dissolved? <blank>13.00</blank>
</answer>

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Answer:

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<reasoning>

Calculate moles of reactants

Using the Molar Mass Calculation and Molarity knowledge points
\[

$$\begin{aligned} \text{Molar mass of NaOH} &= 22.99 + 16.00 + 1.01 = 40.00\text{ g/mol}\\ n_{\text{NaOH}} &= \frac{2.00\text{ g}}{40.00\text{ g/mol}} = 0.0500\text{ mol}\\ n_{\text{HCl}} &= 0.2500\text{ L} \times 0.100\text{ M} = 0.0250\text{ mol} \end{aligned}$$

\]

Determine excess reactant after neutralization

Using the Acid-Base Neutralization Stoichiometry knowledge point
\[

$$\begin{aligned} \text{Reaction:} \quad &\text{NaOH} + \text{HCl} ightarrow \text{NaCl} + \text{H}_2\text{O}\\ n_{\text{NaOH, excess}} &= 0.0500\text{ mol} - 0.0250\text{ mol} = 0.0250\text{ mol} \end{aligned}$$

\]

Calculate hydroxide ion concentration

Using the Molarity knowledge point
\[

$$\begin{aligned} [\text{OH}^-] &= \frac{0.0250\text{ mol}}{0.2500\text{ L}} = 0.100\text{ M} \end{aligned}$$

\]

Calculate pH of the solution

Using the pH and pOH Calculations knowledge point
\[

$$\begin{aligned} \text{pOH} &= -\log_{10}(0.100) = 1.00\\ \text{pH} &= 14.00 - 1.00 = 13.00 \end{aligned}$$

\]
</reasoning>

<answer>
If you added 2.00 grams of solid NaOH to a 250.0 mL solution of 0.100 molar hydrochloric acid, what would be the pH of the resulting solution after all the NaOH had dissolved? <blank>13.00</blank>
</answer>

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