strong acid-strong base systems

what is the concentration of the strong acid?

key
$o$ 0.20 mol $\text{h}^+$
$\blacksquare$ 0.20 mol $\text{a}^-$

5.0 l container

0.24 m acid
0.040 m acid
1.2 m acid

Question

strong acid-strong base systems

what is the concentration of the strong acid?

key
$o$ 0.20 mol $\text{h}^+$
$\blacksquare$ 0.20 mol $\text{a}^-$

5.0 l container

0.24 m acid
0.040 m acid
1.2 m acid

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Acid Strength", "Acid Dissociation Model" ], "new_concepts": [ "Molarity Calculations", "Strong Acid Dissociation" ], "current_concepts": [ "Acid Strength", "Acid Dissociation Model", "Molarity Calculations", "Strong Acid Dissociation" ] } ## Count the particles in the container Using the *Acid Dissociation Model* knowledge point $$ \begin{aligned} &\text{Number of white circles } (\text{H}^+) = 6\ &\text{Number of dark squares } (\text{A}^-) = 6 \end{aligned} $$ ## Calculate total moles of the acid Using the *Acid Strength* knowledge point $$ \begin{aligned} &\text{Since the acid is strong, it dissociates completely: } \text{HA} ightarrow \text{H}^+ + \text{A}^-\ &\text{Moles of } \text{H}^+ = 6 \times 0.20\text{ mol} = 1.2\text{ mol}\ &\text{Moles of initial strong acid } \text{HA} = 1.2\text{ mol} \end{aligned} $$ ## Calculate the molar concentration Using the *Acid Dissociation Model* knowledge point $$ \begin{aligned} &\text{Volume of container} = 5.0\text{ L}\ &\text{Molarity } (M) = \frac{1.2\text{ mol}}{5.0\text{ L}} = 0.24\text{ M} \end{aligned} $$

(A) 0.24 M acid

(B) 0.040 M acid

(C) 1.2 M acid

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Chemistry", "Acid-Base Equilibrium" ] }

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QUESTION IMAGE

Question

Explanation:

Count the particles in the container

Calculate total moles of the acid

Calculate the molar concentration

Answer: