Calculate the pH when 75.20 mL of 0.114 M HCl is titrated with 0 ml of 0.108 M NaOH
The pH at what point?
it doesn't say. That's all that is asked.
Frankly I don't know what it's asking unless by titration they mean the HCl has been exactly neutralized. If that is the case, then the pH will be 7.0 because the salt produced is NaCl and neither the cation nor the anion is hydrolyzed.
The other interpretation is that the zero mL means they want the pH of the HCl at the beginning of the titration. If that is the case you want the pH of the 0.114 M HCl. H^+ = 0.114 and pH = -log(H^+)
To calculate the pH when 75.20 mL of 0.114 M HCl is titrated with 0 mL of 0.108 M NaOH, we need to determine the concentration of the remaining acid after the titration.
Given:
- Initial volume of HCl (V1): 75.20 mL
- Concentration of HCl (C1): 0.114 M
- Volume of NaOH added (V2): 0 mL
- Concentration of NaOH (C2): 0.108 M
Step 1: Determine the number of moles of HCl initially present.
Number of moles of HCl = concentration of HCl (C1) * volume of HCl (V1)
= 0.114 M * 75.20 mL
= 8.5968 mmol
Step 2: Determine the number of moles of NaOH used.
Since no NaOH is added, the number of moles of NaOH used is 0.
Step 3: Determine the number of moles of excess HCl.
Number of moles of excess HCl = moles of HCl initially present - moles of NaOH used
= 8.5968 mmol - 0 mmol
= 8.5968 mmol
Step 4: Determine the volume of excess HCl in liters.
Volume of excess HCl = volume of HCl (V1) - volume of NaOH added (V2)
= 75.20 mL - 0 mL
= 75.20 mL (or convert to L by dividing by 1000)
= 0.07520 L
Step 5: Determine the concentration of the excess HCl in moles per liter (M).
Concentration of excess HCl = moles of excess HCl / volume of excess HCl
= 8.5968 mmol / 0.07520 L
= 114.30 M
Step 6: Calculate the pH of the excess HCl using the formula: pH = -log[H+]
pH = -log[concentration of excess HCl]
= -log(114.30)
≈ 2.94
Therefore, the pH when 75.20 mL of 0.114 M HCl is titrated with 0 mL of 0.108 M NaOH is approximately 2.94.