IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

sachin_gupta1991

Q. Determine the pOH of a solution after 0.1 moles of NaOH is added to 1 L of a solution containing 0.15 M CH₃COOH and 0.2 M CH₃COONa. Assume no change in volume. (K_a=1.8 x 10^-5).

RATES ASSURED TO ALL CORRECT ANSWERS.

Utsav55

0.15 M CH3COOH and 0.2 M CH3COONa
means salt of weak acid and strong base.

And for weak acid and str. base, we have:

pOH = 7 - 1/2( pka + logc )
where c = molar conc. of the salt.
c = 0.1+0.15+0.2 [no. of moles remains constant]

pKa = 1.8*10^-5 [given]

pOH = 7 - 1/2 (1.8*10^-5 + 0.45)
= 7.173

rushi_90

im sorry i dont have time to solve the whole thing right now, but the answer above mine is wrong, ch₃cooh+NaOH==>CH₃COONa find the conc of the above salt +ch3coona already present using the formula, then use ph formula and solve also 0.05 m ch3cooh remains in the solution.

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