Question

100ml of an acid with ph=1 is nixed with 100 ml solution of same acid with ph=2 .The h concn of mixture will be what is the ph of the solution which contains 5% millimoles of sodium formate and 5.8 millimoles of formic acid in 100ml solution. $[ka=1.710^{-4}]$ find

Answer 1

o Determine: The appropriate reagent that must be added to the buffer solution of formic acid and sodium formate to adjust to and also calculate the mass of the added reagent.Answer:To adjust of the buffer solution to strong baseshould be added to the buffer solution.The mass of the strong baseadded to the buffer solution is .Explanation:The volume of solution is.Initial concentration of formic acid is,Initial concentration of sodium formate is,The value of for formic acid is.value is calculated as,The Henderson-Hasselbalch equation is given as,Substitute the values.Therefore initial of the buffer solution is. To increase of the solution to to the solution strong base must be added. Addition of strong base to the buffer solution converts weak acid to its conjugate base and thus increases of the buffer solution.Consider number of moles added to the buffer solution is.Initial mole of formic acid and sodium formate is.Final or new of the buffer solution is.If is added to the buffer solution containing formic acid and sodium formate, acid present in the solution will react with added base and a significant amount of formic acid will be converted to formate ion.The ICE table is as follows, From the ICE table,The Henderson-Hasselbalch equation is given as,Substitute the values.Rearrange and calculate the value of.Therefore the number of moles of added to the buffer solution is .Molar mass of is.Mass of added to the buffer solution is calculated as,



To Determine: The appropriate reagent that must be added to the buffer solution of

formic acid and sodium formate to adjust

pH

to

4.10

and also calculate the mass of the

added reagent.

Answer:

To adjust

pH

of the buffer solution to

4.10

strong base

( )

NaOH

should be added to the

buffer solution.

The mass of the strong base

( )

NaOH

added to the buffer solution is

0.39 g

.

Explanation:

The volume of solution is

250.0 mL

.

Initial concentration of formic acid is,

 

20.025 mol

HCHO 1 L

250.0 mL 1000 mL

0.1 M

=





=

Initial concentration of sodium formate is,

 

20.025 mol

NaCHO 1 L

250.0 mL 1000 mL

0.1 M

=





=

The value of

a

K

for formic acid is

4

1.8 10−



.

a

pK

value is calculated as,

( )

( )

aa

4

p log

log 1.8 10

3.74

KK

−

=−

= − 

=

The Henderson-Hasselbalch equation is given as,

 

2

a

2

CHO

pH p log HCHO

K

−





=+

Substitute the values.

0.1

pH 3.74 log 0.1

3.74

=+

=

Therefore initial

pH

of the buffer solution is

3.74

. To increase

pH

of the solution to

4.10

to the solution strong base must be added. Addition of strong base to the buffer solution

converts weak acid to its conjugate base and thus increases

pH

of the buffer solution.

Consider number of moles

NaOH

added to the buffer solution is

molx

.

Initial mole of formic acid and sodium formate is

0.025 mol

.

Final or new

pH

of the buffer solution is

4.10

.

If

NaOH

is added to the buffer solution containing formic acid and sodium formate, acid

present in the solution will react with added base and a significant amount of formic acid

will be converted to formate ion.

The ICE table is as follows,

( ) ( ) ( )

( )

( )

( )

22

OH HCHO CHO

Before addition mol 0.0 0.025 0.025

Addition mol

After addition mol