To calculate the pH of the resulting solution after mixing HCl and NaOH, we first need to determine the number of moles of HCl and NaOH involved in the reaction.
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Calculate the moles of HCl: \[ \text{Moles of HCl} = \text{Concentration (M)} \times \text{Volume (L)} \] \[ = 0.160 , \text{M} \times 0.0160 , \text{L} = 0.00256 , \text{moles of HCl} \]
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Calculate the moles of NaOH: \[ \text{Moles of NaOH} = \text{Concentration (M)} \times \text{Volume (L)} \] \[ = 0.210 , \text{M} \times 0.0260 , \text{L} = 0.00546 , \text{moles of NaOH} \]
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Determine the reaction: HCl and NaOH react in a 1:1 molar ratio: \[ \text{HCl (aq) + NaOH (aq) } \rightarrow \text{ NaCl (aq) + H}_2\text{O (l)} \]
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Calculate the limiting reagent: We have:
- Moles of HCl = 0.00256
- Moles of NaOH = 0.00546
Since HCl is the limiting reagent, all of it will react with NaOH.
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Calculate the remaining moles of NaOH: \[ \text{Remaining moles of NaOH} = 0.00546 , - , 0.00256 = 0.00290 , \text{moles} \]
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Calculate the total volume of the solution: \[ \text{Total Volume} = 16.0 , \text{ml} + 26.0 , \text{ml} = 42.0 , \text{ml} = 0.0420 , \text{L} \]
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Calculate the concentration of the remaining NaOH: \[ \text{Concentration of NaOH} = \frac{\text{remaining moles}}{\text{total volume (L)}} \] \[ = \frac{0.00290}{0.0420} \approx 0.0690 , \text{M} \]
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Calculate the pOH of the solution: \[ \text{pOH} = -\log[\text{NaOH}] \] \[ = -\log(0.0690) \approx 1.16 \]
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Calculate the pH: \[ \text{pH} + \text{pOH} = 14 \] \[ \text{pH} = 14 - 1.16 \approx 12.84 \]
Thus, the pH of the resulting solution is approximately 12.84.