Question

In: Chemistry

You add 7.50 mL of 0.050 M NaOH to 50.00 mL of pure water, and to...

You add 7.50 mL of 0.050 M NaOH to 50.00 mL of pure water, and to this mixture you then add 1.50 mL of 0.200 HCl. What will be the pH of the resulting solution?

2.11

11.10

11.89

7.00

2.49

Solutions

Expert Solution

Total volume of slution =7.5mL + 50mL + 1.50 mL = 59mL

H+ ions initially = Molarity of HCl solution * Volume of HCl solution = 1.50*0.200 = 0.300 millimoles

OH- ions initially = Molarity of NaOH solution * Volume of NaOH solution = 7.50*0.050 = 0.375 millimoles

OH- and H+ ions will neutralise together in ratio 1:1 and excess OH- ions will be left in solution.

Moles of OH- ions left = 0.375-0.300 = 0.075 millimoles

Conc. of OH- ions[OH-] = No of moles of OH-/ Volume of solution = 0.075/59 = 0.001271M (OH- from water is negligible)

pOH = - log[OH-]

= 2.89

We know pH = 14 - pOH

pH = 14-2.89 = 11.10 (optionB)


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