Question

calculate the required volumes of 0.1M CH3COOH and 0.1M CH3COONa to prepare three 100mL solutions at pH 5.80, 4.75 and 3.50

Answer 1

let x= volume of CH3COOH in L, volume of CH3COONa= 100/1000-x =0.1-x

mole of CH3COOH= 0.1*x, moles of CH3COONa= 0.1*(0.1-x)

[A-]/[HA] = (0.1-x)/x

from pH=pKa+ log [A-]/[HA]

where A- = CH3COO- [HA] =acetic acid, PKa of acetic acid =4.74

5.8= 4,74+ log{[A-]/[HA]}

[A-]/[HA] =11.48

(0.1-x)/x=11.48

0.1-x= 11.48x 0.1 =12.48x, x= 0.1/12.48=0.008L= 8ml and CH3COONa= 100-8 =92ml

b) for pH= 4.75

(0.1-x)/x =10^(4.75-4.74)= 1.023

0.1-x =1.023x

2.023x= 0.1, x =0.1/2.023=0.0494=49,43 and CH3COONa= 100-49.43= 51.57 ml

c) (0.1-x)/x= 10^(3.5-4.74)=0.0575

0.1-x =0.0575x

1.0575x= 0.1 x=0.1/1.0575=0.0945 L= 94.56ml and CH3COONa= 100-94.56= 5.44ml