A 25.00 mL sample of 0.300 M NaOH is titrated with 0.750 M HCl at 25 °C. Calculate the initial pH before any titrant is added. pH = 13.5 Calculate the pH of the solution after 5.00 mL of the titrant is added. pH = 13.8
I'm having issues with the second part of this probem after the addition of 5.00 mL of titrant. I have attached a picture of the problem.
Given : Volume of NaOH solution = 25.00 mL = 0.0250 L (since 1 L = 1000 mL)
Concentration of NaOH = 0.300 M
Concentration of titrant HCl = 0.750 M
And volume of HCl solution added = 5.00 mL = 0.005 L (since 1 L = 1000 mL)
Since moles = concentration X volume of solution in L
=> Moles of HCl added = 0.750 X 0.005 = 0.00375 mol.
And moles of NaOH present = 0.300 X 0.0250 = 0.0075 mol.
The final volume of solution = volume of HCl solution + volume of NaOH solution = 0.005 + 0.0250 = 0.030 L
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