Question

a kapiolani community college student titrated 250.0 ml of a 0.1000 m naoh solution against a solution of h₂so₄. what is the concentration of the h₂so₄ solution if titration required 221.0 ml?
h₂so₄ + 2 naoh → 2 h₃o⁺¹ + na₂so₄
there is enough information to calculate the moles of 1 but not moles of acid
2 mol naoh
(0.2500 l)(-------------) = 3 mol naoh
1 l
4 mol h₂so₃
(5 mol naoh)(-------------) = 6 mol h₂so₄
7 mol naoh
8 mol h₂so₄
------------ = 9 m h₂so₄
10 l
a. acetic b. hydrochloric c. phosphoric d. perchloric e. hydrofluoric

f. sulfuric g. sulfurous h. hydrosulfuric i. oh⁻¹ j. h₃o⁺¹ k. so₄⁻²

l. mg⁺² m. cl⁻¹ n. hpo₃⁻² o. h₂po₃⁻¹ p. co₃⁻² q. hco₃⁻¹

Explanation:

Step1: Calculate moles of NaOH

Molarity formula: \(n = M\times V\). Given \(M = 0.1000\space M\) (moles per liter) and \(V=0.2500\space L\).
\((0.2500\space L)\times\frac{0.1000\space mol\space NaOH}{1\space L}=0.02500\space mol\space NaOH\)

Step2: Use mole - ratio from chemical equation

From \(H_{2}SO_{4}+2NaOH
ightarrow2H_{3}O^{+}+Na_{2}SO_{4}\), mole - ratio of \(H_{2}SO_{4}\) to \(NaOH\) is \(\frac{1\space mol\space H_{2}SO_{4}}{2\space mol\space NaOH}\).
If \(n_{NaOH}=0.02500\space mol\), then \(n_{H_{2}SO_{4}}=(0.02500\space mol\space NaOH)\times\frac{1\space mol\space H_{2}SO_{4}}{2\space mol\space NaOH}=0.01250\space mol\space H_{2}SO_{4}\)

Step3: Calculate molarity of \(H_{2}SO_{4}\)

Molarity formula \(M=\frac{n}{V}\). Given \(n = 0.01250\space mol\) and \(V = 0.2210\space L\) (since \(221.0\space mL=0.2210\space L\)).
\(M=\frac{0.01250\space mol\space H_{2}SO_{4}}{0.2210\space L}\approx0.05656\space M\)

Now for the blanks:

1. \(NaOH\) (because we know volume and molarity of \(NaOH\) initially)
2. \(0.1000\) (molarity of \(NaOH\))
3. \(0.02500\) (\(n = M\times V=0.1000\times0.2500\))
4. \(1\) (from mole - ratio in the reaction \(H_{2}SO_{4}+2NaOH

ightarrow\) products)

1. \(0.02500\) (moles of \(NaOH\) calculated above)
2. \(0.01250\) (\(n_{H_{2}SO_{4}}=\frac{n_{NaOH}}{2}\))
3. \(2\) (mole - ratio of \(NaOH\) to \(H_{2}SO_{4}\) in the reaction)
4. \(0.01250\) (moles of \(H_{2}SO_{4}\))
5. \(0.05656\) (\(M=\frac{n}{V}=\frac{0.01250}{0.2210}\))
6. \(0.2210\) (volume of \(H_{2}SO_{4}\) in liters)

Answer:

1. \(NaOH\)
2. \(0.1000\)
3. \(0.02500\)
4. \(1\)
5. \(0.02500\)
6. \(0.01250\)
7. \(2\)
8. \(0.01250\)
9. \(0.05656\)
10. \(0.2210\)