Answer to Question #138124 in General Chemistry for BRIANNA SMITH

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