Question

find the ph of each mixture of acids

0.020m HNO3 and 0.225 Mbr

0.095m in acetic acid and 0.075.m in H2SO4
0.0500 M in HCL04 and 0.02 M HBR

Answer 1

find the ph of each mixture of acids

0.020m HNO3 and 0.225 M HBr

0.095m in acetic acid and 0.075.m in H2SO4
0.0500 M in HCL04 and 0.02 M HBR

1.

0.020 M HNO3 and 0.225 M HBr

0.020 M HNO3 = 0.020 [H+]

0.225 M HBr = 0.225 [H]

Thus there are total; [H+]= 0.020+0.225

=0.245

pH = -log [H+]

= -log 0.245

= 0.61

2.

0.095 M in acetic acid and 0.075. M in H2SO4

CH3COOH(aq) + H2O(l) -----> CH3COO-(aq) + H3O+(aq)

Ka is acid dissociation constant; 1.74 * 10^-5

Ka= [CH3COO-] * [H3O+] / [CH3COOH]

1.74 * 10^-5 = x*x /0.095 M-x
here Ka is very less than 0.095 M-x =0.095 M

1.74 * 10^-5 = [x]^2 / 0.095
[x]^2 = 1.65*10^-6

[x]= 1.28*10^-3 = [H+] from acetic acid

[H+] from H2SO4 = 0.075*2= 0.15

Total [H+]= 1.28*10^-3+0.15

= 0.15128

pH = -log [H+]

= -log 0.15128

= 0.82

3.

0.0500 M in HCL04 and 0.02 M HBr

[H+] from HCL04 = 0.0500

[H+] from HBr = 0.02

Total [H+]= 0.02++0.0500

= 0.07

pH = -log [H+]

= -log 0.07

= 1.15