In: Chemistry
What is the pH of the 25 mL of 0.200 M acetic acid when you add
70 mL of 0.1M NaOH
pKa = 4.75
no of moles of CH3COOH = molarity * volume in L
= 0.2*0.025 = 0.005 moles
no of moles of NaOH = molarity * volume in L
=0.1*0.07 = 0.007 moles
no of mole of NaOH excess = 0.007-0.005 = 0.002 moles
total volume = 25+70 = 95ml = 0.095L
conc of NaOH = no of moles/volume in L
= 0.002/0.095 = 0.021 M
NaOH --------> Na^+ (aq) + OH^- (aq)
0.021M 0.021M
[OH^-] = [NaOH]
[OH^-] = 0.021M
POH = -log[OH^-]
= -log0.021 = 1.6777
PH = 14-POH
= 14-1.6777 = 12.3223 >>>>answer