Questions: If 10.0 mL of 0.20 M NaOH is added to 50.0 mL of 0.10 M HCl, what will be the pH of the resulting solution? - Round your answer to two decimal places. Provide your answer below: pH=

Solution Steps

First, calculate the moles of NaOH and HCl in the solutions.

For NaOH: \[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.010 \, \text{L} \times 0.20 \, \text{M} = 0.0020 \, \text{mol} \]

For HCl: \[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.050 \, \text{L} \times 0.10 \, \text{M} = 0.0050 \, \text{mol} \]

NaOH and HCl react in a 1:1 molar ratio: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

Since there are 0.0020 mol of NaOH and 0.0050 mol of HCl, NaOH is the limiting reactant. All 0.0020 mol of NaOH will react with 0.0020 mol of HCl, leaving: \[ \text{Remaining moles of HCl} = 0.0050 \, \text{mol} - 0.0020 \, \text{mol} = 0.0030 \, \text{mol} \]

The total volume of the solution after mixing is: \[ \text{Total Volume} = 10.0 \, \text{mL} + 50.0 \, \text{mL} = 60.0 \, \text{mL} = 0.060 \, \text{L} \]

The concentration of the remaining HCl is: \[ \text{Concentration of HCl} = \frac{\text{Remaining moles of HCl}}{\text{Total Volume}} = \frac{0.0030 \, \text{mol}}{0.060 \, \text{L}} = 0.050 \, \text{M} \]

Since HCl is a strong acid, it dissociates completely. The concentration of \( \text{H}^+ \) ions is equal to the concentration of HCl: \[ [\text{H}^+] = 0.050 \, \text{M} \]

The pH is calculated using the formula: \[ \text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(0.050) \]

Calculating the pH: \[ \text{pH} \approx 1.30 \]

Final Answer