2) Calculate the pH of a buffer solution that contains 0.20 M benzoic acid (C6H5

Question

2) Calculate the pH of a buffer solution that contains 0.20 M benzoic acid (C6H5CO2H) and 0.15 M sodium benzoate (C6H5COONa). [Ka = 6.5 x 10–5 for benzoic acid]

3) You are asked to go into the lab and prepare an acetic acid - sodium acetate buffer solution with a pH of 4.00. What molar ratio of CH3COOH to CH3COONa should be used? (Ka of acetic acid is 1.8x10-5)

4)You have 500.0 mL of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the pH of this solution be after the addition of 20.0 mL of 0.80 M NaOH solution? [Ka = 1.8 x10–5]

Explanation / Answer

2 ) pH of acidc buffer = pka + log(salt/acid)

       pka of c6h5cooh = -log(6.5*10^(-5)) = 4.12

= 4.12+log(0.15/0.2)

= 3.99

3) pka of ch3cooh = -log(1.8*10^(-5)) = 4.74

pH of acidc buffer = pka + log(salt/acid)

                    4 = 4.74 + log(CH3COONa/CH3COOH)

    (CH3COONa/CH3COOH)   = 0.18

(CH3COOH/CH3COONa) = 1/0.18 = 5.55

4) No of moles of CH3COOH = 500/1000*0.3 = 0.15 mole

Noof moles of CH3COONa = 500/1000*0.2 = 0.1 mole

No of moles of NaOH added = 20/1000*0.8 = 0.016 mole

= 4.74+log((0.1+0.016)/(0.15-0.016))

= 4.68

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.