1. Prepare your first buffer by mixing 50.0 mL of 0.100 M NaOAc and 45.0 mL of 0

Question

1. Prepare your first buffer by mixing 50.0 mL of 0.100 M NaOAc and 45.0 mL of 0.100 M HOAc. Use graduated cylinders for these volume measurements. Mix well.

2. Prepare your second buffer by mixing 50.0 mL of 0.100 M NaOH and 95.0 mL of 0.100 M HOAc (or, if available, 47.5 mL of 0.200 M HOAc). Use graduated cylinders for these volume measurements. Mix well.

Considering the molar concentration of acetic acid, what are the buffer mol/L values for mixtures 1 and 2?

Considering the molar concentration of acetate ion, what are the buffer mol/L values for mixtures 1 and 2?

Considering that pH=pKa+log{[B-]/[HB]}, what will the buffer pH value be for mixtures 1 and 2?

Explanation / Answer

In Buffer 1

50.0 mL of 0.100 M NaOAc and 45.0 mL of 0.100 M HOAc is present

N1V1 =N2V2

[NaOAc] X V1 = N2 X 95; N2 = 0.052

[HOAc] X V1 = N2 X 95 ; N2 = 0.047

hence

pH will be

pH = pKa + log ([salt] / [acid])

pH = 4.75 + log (0.052/0.047)

pH = 4.79

Buffer 2

since some amount of [HOAc] is consumed to become [HOAc] ;

[HOAc] = 95 X 0.1 = 9.5 meq

[NaOH] = 50 X 0.1

amount of acid left after mixing =

9.5-5=4.5 meq

or 4.5/145; = 0.031M

hence amount of [HOAc] formed = amount of [NaOH] consumed

5/145 = 0.034 M

pH = pKa + log ([salt] / [acid])

pH = 4.75 + log (0.034/0.031)

pH = 4.79

Considering the molar concentration of acetic acid, the buffer mol/L values for mixtures 1 is 0.099 and Buffer2 is 0.065M/L

Considering the molar concentration of acetate, the buffer mol/L values for mixtures 1 is 0.099 and Buffer2 is 0.065M/L

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