HCl moles = M x V ( inL) = 0.25 x 0.025 = 0.00625 = H+ moles NaOH moles = 0.25 x
ID: 699617 • Letter: H
Question
HCl moles = M x V ( inL) = 0.25 x 0.025 = 0.00625 = H+ moles
NaOH moles = 0.25 x 0.03 = 0.0075 = OH- moles
we have reatcion H+ (aq ) + OH- (aq) <--> H2O (l)
OH- moles left after reaction = 0.0075-0.00625 = 0.00125
solution volume = 0.025+0.03 =0.055 L
[OH-] = moles / volume = 0.00125 /0.055 = 0.0227 M
pOH = -log [OH-] = -log ( 0.0227) = 1.64
pH = 14-1.64 = 12.36
b) OH- moles = 0.3 x 0.035 = 0.0105
OH- moles left after reacting with H+ = 0.0105 - 0.00625 = 0.00425
solution volume = 35ml + 25ml = 60 ml
[OH-] = 0.00425 / 0.06 = 0.07083 M
pOH = -log [OH-] = - log ( 0.07083) = 1.15
pH = 14-1.15 = 12.85
Explanation / Answer
Image Nat Found Calculate the pH of the resulting solution if 25.0 mL of 0.250 M HCI(aq) is added to (a) 30.0 mL of 0.250 M NaOH(aq). Number (b) 35.0 mL of 0.300 M NaOH(aq). Number contact
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