Question

75.0ml of 1.00M HCl is added to 100ml of 1.00M NaOH. What is the pH of the resulting solution? (Please show your work.)

Answer 1

75.0ml of 1.00M HCl is added to 100ml of 1.00M NaOH

mols of HCl present = 75 ml * 1 M = 75 milli mols

mols of NaOH present = 100 ml * 1 M = 100 milli mols

Since both acid and base are strong, and they react in 1:1 mol ratio,

75 milli mols of HCl will react with 75 milli mols of NaOH

Thus there will be , 100 - 75 millimol = 25 millimol NaOH remains in the

solution. 

Thus [NaOH] = 25 m mol/(100 +75 ml) = 0.14285 M

NaOH--> Na + + OH-

thus [OH-] = 0.14285 M

pOH = -log [OH-] = 0.845

pH = 14-pOH = 13.15

ANSWER

pH = 13.2