2. What is the pH of the buffer created by dissolving 60.0 g HCN (MM =...

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Question

Answer 1

Answer #1

Q.NO1.Pka=-logka=-log(4X10-10)

=10log1010-log4=10-0.602=9.398 (log1010=1)

concentration of HCN=W/MM*1000/V=60/27.053*1000/750=2.949M

CONCENTRATION OF KCN=W/MM*1000/V=60/65.12*1000/750=1.22M

PH=Pka +log[salt]/[acid]

=9.398+log[1.22]/2.949

=9.398+log(0.413)=9.398+0.384=9.782

Q.NO.2 AFTER ADDING 15 ml HCL FIND PH

CONCENTRATION OF Hcl =w/mm*1000/v=15/36.61*1000/750=0.547

PH=Pka +log[salt]/[acid]

Pka=9.398

pH=9.398+log[1.22]/0.547=9.398+log(2.2)

=9.398+0.342=9.74

after adding Hcl pH value decreases

Answer 2

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