Question

30.0 mL of 0.512 M sulfuric acid was added to 25.0 mL of 0.666 M sodium

hydroxide.

a. Write the **balanced chemical equation**.

b. Determine the **molarity** of **either
acid** or **base** remaining after the
reaction.

c. Find the **molarity** if a 50.00 mL aliquot of
this solution is diluted to 250. mL.

Answer #1

Given,

Concentration of H_{2}SO_{4} solution =
**0.512 M**

Volume of H_{2}SO_{4} solution = 30.0 mL x (
1L/1000 mL) = **0.03 L**

Concentration of NaOH solution = **0.666 M**

Volume of NaOH solution = 25.0 mL x ( 1L/1000 mL) =
**0.025 L**

**a)** The balanced chemical reaction between
H_{2}SO_{4} and NaOH is,

**H _{2}SO_{4}(aq) + 2NaOH(aq)
Na_{2}SO_{4}(aq) + 2H_{2}O(l)**

**b)** Now, Calculating the number of moles of
sulfuric acid and sodium hydroxide given,

= 0.512 M x 0.030 L = **0.0154 mol
H _{2}SO_{4}**

Similarly,

= 0.666 M x 0.025 L = **0.01665 mol NaOH**

Now, Calculating the number of moles of NaOH required to react
completely with the given moles of H_{2}SO_{4},

= 0.0154 mol H_{2}SO_{4} x ( 2 mol NaOH / 1 mol
H_{2}SO_{4})

**= 0.03072 mol NaOH is required to react completely with
the given moles of H _{2}SO_{4}.**

Also, Calculating the number of moles of
H_{2}SO_{4} required to react completely with the
given moles of NaOH,

= 0.01665 mol NaOH x ( 1 mol H_{2}SO_{4} / 2 mol
NaOH)

**= 0.008325 mol H _{2}SO_{4} is required
to react completely with the given moles of NaOH.**

Thus, NaOH is limiting reactant and H_{2}SO_{4}
is an excess reactant.

Now, calculating the excess reactant,

= Given moles of H_{2}SO_{4} - Moles of
H_{2}SO_{4} reacted.

= 0.0154 mol H_{2}SO_{4} - 0.008325 mol

**= 0.007035 mol of
H _{2}SO_{4}**

Now,

the total volume = 0.03 L + 0.025 L

**total volume = 0.055 L**

New molarity(M) = Number of moles / Volume of solution

New molarity(M) = 0.007035 mol of H_{2}SO_{4} /
0.055 L solution

**New molarity(M) = 0.128 M**

**c)** We know,

Molarity of solution = **0.128 M**

Volume of the aliquot of this solution taken(V) = 50.00 mL x (
1L /1000 mL) = **0.05 L**

Calculating the number of moles

= 0.128 M x 0.05 L

= 0.006395 moles

Now, dilution volume = 250 mL /1000 mL = **0.250
L**

**Molarity = Moles / L of solution**

Molarity = 0.006395 moles / 0.250 L of solution

**Molarity = 0.0256 M**

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