What is the ph of a sodium formate solution prepared by adding 0.680 grams of sodium formate to 100.0 ml of water at 25.0°c? the ka at 25.0 °c for formic acid is 1.8 x 10-4. enter your answer with two decimal places.

Answer : The pH of the solution is, 8.38

Solution :  Given,

Concentration (c) = 0.11 M

Acid dissociation constant =

First we have to calculate the moles of sodium formate.

Now we have to calculate the concentration of formic acid.

Concentration of formic acid = Concentration of sodium formate =

Now we have to calculate the base dissociation constant.

The equilibrium reaction for dissociation of (weak base) is,

initially conc.         0.1                             0             0

At eqm.              (0.1-x)                            x             x  

First we have to calculate the value of 'x'.

Formula used :

Now put all the given values in this formula ,we get:

By solving the terms, we get

Thus, the concentration of hydroxide ion is:

Now we have to calculate the pOH.

Now we have to calculate the pH.

Therefore, the pH of the solution is, 8.38

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