Let's determine the molarity of the hydrochloric acid (HCl) when it is neutralized by ammonia (NH₃). Here's a detailed step-by-step solution:
### Step 1: Write the balanced chemical equation
The reaction between ammonia (NH₃) and hydrochloric acid (HCl) results in the formation of ammonium chloride (NH₄Cl):
[tex]\[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl} \][/tex]
This shows a 1:1 molar ratio between NH₃ and HCl.
### Step 2: Determine the moles of NH₃
First, calculate the moles of NH₃ using its molarity and volume:
[tex]\[
\text{Molarity of NH}_3 = 0.75 \, \text{M}
\][/tex]
[tex]\[
\text{Volume of NH}_3 = 19.5 \, \text{mL} = 19.5 \times 10^{-3} \, \text{L}
\][/tex]
[tex]\[
\text{Moles of NH}_3 = \text{Molarity} \times \text{Volume} = 0.75 \, \text{M} \times 19.5 \times 10^{-3} \, \text{L}
\][/tex]
[tex]\[
\text{Moles of NH}_3 = 0.75 \times 19.5 \times 10^{-3}
\][/tex]
[tex]\[
\text{Moles of NH}_3 = 14.625 \times 10^{-3} \, \text{mol}
\][/tex]
### Step 3: Determine the moles of HCl
From the balanced chemical equation, we know that the moles of HCl will be equal to the moles of NH₃ because the molar ratio is 1:1:
[tex]\[
\text{Moles of HCl} = \text{Moles of NH}_3 = 14.625 \, \text{mmol} = 14.625 \times 10^{-3} \, \text{mol}
\][/tex]
### Step 4: Calculate the molarity of HCl
We need to find the molarity (M) of HCl using its volume and the moles:
[tex]\[
\text{Volume of HCl} = 43.0 \, \text{mL} = 43.0 \times 10^{-3} \, \text{L}
\][/tex]
[tex]\[
\text{Molarity of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl in liters}}
\][/tex]
[tex]\[
\text{Molarity of HCl} = \frac{14.625 \times 10^{-3} \, \text{mol}}{43.0 \times 10^{-3} \, \text{L}}
\][/tex]
[tex]\[
\text{Molarity of HCl} = \frac{14.625}{43.0}
\][/tex]
[tex]\[
\text{Molarity of HCl} \approx 0.34 \, \text{M}
\][/tex]
Therefore, the molarity of the hydrochloric acid (HCl) is approximately 0.34 M.
