Question

A 1.0-l buffer solution contains 0.100 mol hc2h3o2 and 0.100 mol nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. calculate the ph of the solution upon addition of 58.9 ml of 1.00 m hcl to the original buffer.

Answer 1

58.9 mL of 1 M HCl added to the buffer,

So the moles of HCl=\frac{58.9\times 1}{1000}

=0.058 mol

Now as more HCl is added, it will combine with acetate ion to form acetic acid.

CH₃COO⁻ + H⁺ ----->CH₃COOH

So new buffer contains,

0.1+0.058=0.158 moles of acetic acid (CH3COOH)

0.1 - 0.01 = 0.09 moles of CH3COO⁻

As K_{a}=1.8\times 10^{-5}

pK_{a}= -log(1.8\times 10^{-5})

Now pH can be calculated using equation;

pH=pK_{a}+log \frac{[CH_{3}COO^{-1}]}{[CH_{3}COOH]}

putting all the values in the equation;

pH=-log(1.8\times 10^{-5})+log \frac{[0.158]}{[0.09]}

pH=4.93.

Answer 2

58.9 mL of 1 M HCl added to the buffer,

So the moles of HCl=\frac{58.9\times 1}{1000}

=0.058 mol

Now as more HCl is added, it will combine with acetate ion to form acetic acid.

CH₃COO⁻ + H⁺ ----->CH₃COOH

So new buffer contains,

0.1+0.058=0.158 moles of acetic acid (CH3COOH)

0.1 - 0.01 = 0.09 moles of CH3COO⁻

As K_{a}=1.8\times 10^{-5}

pK_{a}= -log(1.8\times 10^{-5})

Now pH can be calculated using equation;

pH=pK_{a}+log \frac{[CH_{3}COO^{-1}]}{[CH_{3}COOH]}

putting all the values in the equation;

pH=-log(1.8\times 10^{-5})+log \frac{[0.158]}{[0.09]}

pH=4.93.

Answer 3

RKGIOJREITG90954389289 IS THE ASNER (b)