NaOH(09). CH3COOH.ag) -H2OM-CH3COONa cap Sodium actic -. Water . nYOYondi aaa au
ID: 511102 • Letter: N
Question
NaOH(09). CH3COOH.ag) -H2OM-CH3COONa cap Sodium actic -. Water . nYOYondi aaa autak pHz 4 5 Pk A; Line aco ic acid Ka 1.74 y IO-s NaOH : Na-o-H --- Water 0 01 HM Acetic : H-c-c-6H Aaton Sodium ped .Acetate - H-C-C-ONa ph Pea . Iog facidT- Haan /00 [con] base) raad 4.5-476 log [mg] 6-109 [base acid] .26 : Ebo sez JC acid 10-,2G= 155" [base 55faciaTLaciaJ -B: 2.3c ka [HOT B: 2.3c iklsoorl' lap H2O+3] 2 6.74-10-5][3.16 x10" 5] 5.53x10"to .229 [1A -10-s 13, 16 xio-s32 214)xIO-9 ,53c ICs, 189 c, [samfociaJ C,189 C= [SaMHaciaJExplanation / Answer
Acetic acid/acetate buffer will be a proper choice.
pH = Pka + log [acetate/acetic acid]
4.5 = 4.76 + log [acetate/acetic acid]
[acetate/acetic acid] = 0.55
Calculate the decimal fraction :
[acetate] = 0.55/1+0.55 = 0.35
[acetic acid ] = 1/1+0.55 =0.65
Suppose you want 1M and 0.1L buffer solution.
Molarity of acetate = 1M * 0.35 = 0.35 M
Moles of acetate = 0.35 M * 0.1L = 0.035 moles
Mass of sodium acetate = 0.035 moles * 82g/mol =2.87 gm
Molarity of acetic acid = 0.65 * 1M = 0.65 M
Moles of acetic acid = 0.65 M * 0.1 L = 0.065 moles
Mass of acetc acid = 0.065 moles * 60gm/mol = 3.9 gm
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If 1mL 1 M HNO3 is added to the buffer, H+ from the acid will react with acetate to convert it to acetic acid. So, the concentration of acetate will increase and acetic acid will decrease in the solution.
Moles of acid added = 1M * 0.001 L = 0.001 moles
Moles of acetic acid present after adding HNO3 = 0.065 + 0.001 = 0.066 moles
Moles of acetate present after adding HNO3 = 0.035 - 0.001 = 0.034 moles
pH = pKa + log [acetate/acetic acid]
= 4.74 + log [0.034/0.066] = 4.45
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when 1 mL 1M NaOh is added, it will react with acetic acid to convert it to acetate. So that concentration of acetate increases in the solution and acetic acid decreases in the solution.
Moles of base added = 0.001 moles
moles of acetic acid present after adding NaOH = 0.065 -0.001 = 0.064 moles
moles of acetate present after adding NaOH = 0.035+0.001 = 0.036 moles
pH = 4.74 + log [0.034/0.064] = 4.46
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