A 40.0 mL solution of a weak base, BOH is titrated with 0.1 N HCl solution. The pH of the solution is found to be 10.04 and 9.14 after adding 5.0 mL and 20.0 mL of the acid respectively. The dissociation constant of the base is:

Q. A solution of weak acid

H A

was titrated with base

N a O H

. The equivalence point was reached when

36.12 m L

0.1 M

N a O H

has been added. Now

18.06 m L

0.1 M

H C l

were added to titrated solution, the

p H

was found to be

4.92

. What will be the

p H

of the solution obtained by mixing

10 m L

0.2 M

N a O H

and

10 m L

0.2 M

H A

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The equivalent point in a titration of 40.0 mL of a solution of a weak monoprotic acid occurs when 35.0 mL of a 0.10 M NaOH solution has been added. The pH of the solution is 5.75 after the addition of 20.0 mL of NaOH solution. What is the dissociation constant of the acid?

Va=40ml, Vb=35ml, NaOH molarity=0.1M pH=5.5H1V1=M2V2M1=0.1×3540=0.0875MpH=−log[H+][H+]=10−45[H+]=3.16×10−6HA⇌H++A−Ka=[H+][A−][HA][H+]=[A−]Ka=(3.16×10−6)20.0875Ka=1.1×10−10Dissociation Constant of acid=1.1×10−10

The equivalent point in a titration of $40.0 m L$ of a solution of a weak monoprotic acid occurs when $35.0 m L$ of a $0.10 M N a O H$ solution has been added. The $p H$ of the solution is $5.75$ after the addition of $20.0 m L$ of $N a O H$ solution. What is the dissociation constant of the acid?