Calculate the buffer capacity of a buffer solution which is prepared by mixing 10ml 0.01M formic acid and 100ml 0.001M sodium formate if Ka for formic acid Is 1 * 10^-3?
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I don't know exact solution but hope this solution helps you...
(a) Calculate the buffer capacity
Buffer capacity is the amount of a monoprotic strong acid or base that must be added to 1 L of a buffer to change its pH by one unit.
Let's assume that we are adding NaOH to the buffer.
The base will decrease the amount of acetic acid and increase the amount of acetate.
Then we have
mmmmmlHA + H2O⇌H3O++A-
I/mol:mll0.01mmmmmmmmmll0.01
C/mol:mll-xmmmmmmmmmmll+x
E/mol:m0.01-xmmmmmmmml0.01+x
pKa=4.76
The pH will increase to 5.76.
pH=pKa+log([A-][HA])
5.76=4.76+log(0.01+x0.01−x)
1.00=log(0.01+x0.01−x)
0.01+x0.01−x=101.00=10.0
0.01+x=10.0(0.01−x)=0.10−10.0x
11.0x=0.09
x=0.0911.0=0.008
The buffer capacity is 0.008 mol.
(b) Calculate the change in pH on adding base
mmmmmmllHA + H2O⇌H3O++A-
I/mol⋅L-1:ml0.01mmmmmmmmml0.01
C/mol⋅L-1:ll-0.005mmmmmmmll+0.005
E/mol⋅L-1:m0.005mmmmmmmmll0.015
For the original buffer,
pH=pKa+log([A-][HA])=4.76+log(0.10.1)=4.76+log1=4.76+0=4.76
After adding the base,
pH=pKa+log([A-][HA])
pH=4.76+log(0.0150.005)=4.76+log3=4.76+0.48=5.24
ΔpH=5.24−4.76=0.48
Hope this helps you....
Answer:
hey mate..........,
(a) Calculate the buffer capacity
Buffer capacity is the amount of a monoprotic strong acid or base that must be added to 1 L of a buffer to change its pH by one unit.
Let's assume that we are adding NaOH to the buffer.
The base will decrease the amount of acetic acid and increase the amount of acetate.
Then we have
mmmmmlHA + H2O⇌H3O++A-
I/mol:mll0.01mmmmmmmmmll0.01
C/mol:mll-xmmmmmmmmmmll+x
E/mol:m0.01-xmmmmmmmml0.01+x
pKa=4.76
The pH will increase to 5.76.
pH=pKa+log([A-][HA])
5.76=4.76+log(0.01+x0.01−x)
1.00=log(0.01+x0.01−x)
0.01+x0.01−x=101.00=10.0
0.01+x=10.0(0.01−x)=0.10−10.0x
11.0x=0.09
x=0.0911.0=0.008
The buffer capacity is 0.008 mol.
(b) Calculate the change in pH on adding base
mmmmmmllHA + H2O⇌H3O++A-
I/mol⋅L-1:ml0.01mmmmmmmmml0.01
C/mol⋅L-1:ll-0.005mmmmmmmll+0.005
E/mol⋅L-1:m0.005mmmmmmmmll0.015
For the original buffer,
pH=pKa+log([A-][HA])=4.76+log(0.10.1)=4.76+log1=4.76+0=4.76
After adding the base,
pH=pKa+log([A-][HA])
pH=4.76+log(0.0150.005)=4.76+log3=4.76+0.48=5.24
ΔpH=5.24−4.76=0.48
hope it's help.....
mark as brainliest......