Question

How many milliliters of \(0.2315 \mathrm{M}\) sulfuric acid are required to neutralize \(38.00 \mathrm{~mL}\) of \(0.189 \mathrm{M}\) ammonia?

Short answer

Answer: 15.50 mL of 0.2315 M sulfuric acid is required to neutralize 38.00 mL of 0.189 M ammonia.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and ammonia (NH3). \[2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4\] The balanced equation tells us that 2 moles of ammonia react with 1 mole of sulfuric acid to form 1 mole of ammonium sulfate.

Find the moles of ammonia

Now, we need to find how many moles of ammonia are present in the given 38.00 mL solution. Use the molarity formula: Moles = Molarity × Volume (in liters) Moles of NH3 = \(\text{Molarity}_{NH_3}\) × \(\frac{\text{Volume}_{NH_3}}{1000}\) Moles of NH3 = \(0.189 \times \frac{38.00}{1000} = 0.007182 \text{ moles}\)

Determine moles of H2SO4 required

Using the balanced chemical equation, we can find the moles of sulfuric acid needed to neutralize the given moles of ammonia. \(2\text{ moles }\text{NH}_3 : 1\text{ mole }\text{H}_2\text{SO}_4\) So, for 0.007182 moles of NH3: Moles of H2SO4 = \(\frac{1}{2} \times 0.007182 = 0.003591\text{ moles}\)

Calculate the volume of sulfuric acid required

Now, we need to find the volume of 0.2315 M sulfuric acid containing 0.003591 moles. Use the molarity formula again, rearranged for volume: Volume (in liters) = \(\frac{\text{Moles}}{\text{Molarity}}\) \(\text{Volume}_{H_{2}SO_{4}}\) (in liters) = \(\frac{0.003591}{0.2315} = 0.01550\text{ L}\) Convert the volume to milliliters: \(\text{Volume}_{H_{2}SO_{4}}\) (in mL) = \(0.01550 \times 1000 = 15.50\text{ mL}\)

State the final answer

To neutralize 38.00 mL of 0.189 M ammonia, 15.50 mL of 0.2315 M sulfuric acid is required.