Pre-Lab calculating pH and Buffer Capacity You must show work for credit) Name:

Question

Pre-Lab calculating pH and Buffer Capacity You must show work for credit) Name: 1. calculate the volume orglacial acetic acia 176 Mo nd mass of sodium (this question is to make 102 ml of 0.20 M buffer at pH 485. The Ka of acetic acid is 1.4x10" and your book for help. challenging, use the background 8203 2. Consult the table of values the back of your textbook and suggest acid conjugate base system only) would yield a strong capacity of 8.60 buffer must be strong compared to others of the same molarity) Explain selection (simply and succinctly) 3 A buffer solution is prepared by mixing 20.0 mL of0.032 M benzoic acid CHscooHu solution with 25.0 mL of 0.022 M Nac Hscoo- What is the pH of the buffer produced?(K. of benzoic acid 6.2

Explanation / Answer

Acetate buffer composition:

volume = 102 mL

pH = 4.85

molarity = 0.2 M

pH = pKa + log (CH3COO- CH3COOH)

Ka = 1.4*10-5

pKa = - log Ka = - log  1.4*10-5 = 4.85

since we get pH = pKa;

Thus,  log (CH3COO- CH3COOH) = 0

(CH3COO- CH3COOH) = 100 = 1

CH3COO- = CH3COOH

since, buffer to be prepared is 0.2 M ,102 mL

[CH3COO- ] + [CH3COOH] = 0.2 M

2 [CH3COOH]= 0.2 M { since CH3COO- = CH3COOH}

[CH3COOH] = 0.1 M

[CH3COO- ] = 0.1 M

firstly we will solve for acetic acid,

17.6 M *V1 = 0.1 M * 102 mL

V1 = 0.57 mL

volume of acetic acid require is 0.57 mL

Now to determine the amount of sodium acetate required;

no. of moles = Molarity * volume of solution in L

no. of moles = 0.1 M * 0.102 L = 0.0102 moles

no. of moles = weight molecular weight

0.0102 = weight 82.03 g

weight = 0.8367 g of sodium acetate is required.

3.

pH = pka + log (benzoateenzoic acid)

Ka = 6.2 *10 -5

pKa = - log Ka = - log 6.2 *10 -5 = 4.2

[ benzoate] = no. of moles = molarity * volume of solution in L = 0.022*25@0 = 0.0005 moles

[benzoic acid] = no. of moles = molarity * volume of solution in L = 0.032*20@0 =0.0006 moles

pH = pka + log (benzoateenzoic acid)

= 4.2 + log[0.0005 moles 0.0006 moles] = 4.12

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