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.

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.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Molar Mass Calculation", "Molarity", "Acid-Base Neutralization Stoichiometry", "pH and pOH Calculations" ], "new_concepts": [], "current_concepts": [ "Molar Mass Calculation", "Molarity", "Acid-Base Neutralization Stoichiometry", "pH and pOH Calculations" ] } ## 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 $$ \begin{aligned} \text{pOH} &= -\log_{10}(0.100) = 1.00\ \text{pH} &= 14.00 - 1.00 = 13.00 \end{aligned} $$ 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? 13.00 { "subject": "Natural Science", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Natural Science", "Chemistry", "Acid-Base Stoichiometry and pH" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate moles of reactants

Determine excess reactant after neutralization

Calculate hydroxide ion concentration

Calculate pH of the solution

Answer: