In: Chemistry
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
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)