To determine the pH of the solution after adding 15.0 mL of 0.100 M NaOH to 20.0 mL of 0.100 M HCl, we first need to figure out how many moles of HCl and NaOH are present.
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Calculate moles of HCl: \[ \text{Moles of HCl} = \text{Volume (L)} \times \text{Concentration (M)} = 0.0200 , \text{L} \times 0.100 , \text{M} = 0.00200 , \text{mol} \]
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Calculate moles of NaOH added: \[ \text{Moles of NaOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.0150 , \text{L} \times 0.100 , \text{M} = 0.00150 , \text{mol} \]
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Determine the reaction and the remaining moles after reaction: The neutralization reaction is: \[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
After 15.0 mL of NaOH is added, the moles of HCl and NaOH will react in a 1:1 molar ratio:
- Initially, we have 0.00200 mol of HCl and add 0.00150 mol of NaOH.
- Moles of HCl remaining after reaction: \[ 0.00200 , \text{mol HCl} - 0.00150 , \text{mol NaOH} = 0.00050 , \text{mol HCl} \]
- Moles of NaOH remaining: Since all of the NaOH reacts, the moles of NaOH left is: \[ 0.00150 , \text{mol NaOH} - 0.00150 , \text{mol NaOH} = 0 , \text{mol} \]
Since there is still HCl left in the solution, the solution is acidic. We will calculate the concentration of the remaining HCl.
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Calculate the total volume of the solution: Total volume after addition of NaOH: \[ 20.0 , \text{mL HCl} + 15.0 , \text{mL NaOH} = 35.0 , \text{mL} = 0.0350 , \text{L} \]
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Calculate concentration of the remaining HCl: \[ \text{Concentration of HCl} = \frac{0.00050 , \text{mol}}{0.0350 , \text{L}} \approx 0.01429 , \text{M} \]
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Calculate the pH of the solution: For a strong acid like HCl, the pH can be calculated directly from the concentration of HCl: \[ \text{pH} = -\log[H^+] = -\log(0.01429) \approx 1.84 \]
Thus, the pH of the resulting solution after adding 15.0 mL of NaOH to the HCl solution is approximately 1.84.