Physics, Chemistry, Maths Forum

Ionic Equilibrium!!!

Ionic Equilibrium!!!

by Asif Zafar -
Number of replies: 2
  • The concentration of hydrogen ion in a 0.2 M solution of formic acid is 6.4 * 10-3 moles per litre. To this solution sodium formate is added so as to adjust the concentration of sodium formate one mole per litre. What will be the pH of this solution? The dissociation constant of formic acid is 2.4 * 10-4 and the degree of dissociation of sodium formate is 0.75

  • 2.05g of sodium acetate was added to 100 ml of .1 M HCl solution. Find the H+ concentration of the resulting solution. If 6 ml of 1M HCl is further added to it, what will be the new H+ concentration? The dissociation constant of acetic acid is 1.8 * 10-3.

Plz give solutions with steps. The 1st question was asked at IITJEE1985. the second is from thhe unsolved questions list in R.C. Mukerjee

In reply to Asif Zafar

Re: Ionic Equilibrium!!!

by karthik vs -

Hi asif-

There are a mistakes in the second qn correcting which will give u the answer to both the parts!!

firstly  the dissociation constant is not 1.8 * 10-3 but it is 1.8 * 10-5 which is usually given in all the problems but was missing in the book!

the second is that the 6ml HCl added is not 1M but it is 0.1M !!!

making these both corrections in the question u will get both the answers accurately!!!

i just used the formula:

Ph= pka + log Undetermined error: @@[salt]/[acid]@@

which gives

ph= 4.774 + log Undetermined error: @@ 0.015/0.1 @@

   = 4.91

[H+]  =  10 -4.91   = 1.23 * 10-5 M

the same formula is applied to solve the second part also

in this case the [salt]=0.0144M and [acid]= 0.0106M.

Hope u get it!!kvs

In reply to Asif Zafar

Re: Ionic Equilibrium!!!

by Shashank Todwal -

sodium formate being a solid powder does not affect volume of solution.

Let conc of sod.formate on just adding it to the solution be C1.After that it is dissociated 75%.So

HCOONa-----><-----HCOO-   + Na+

C1-0.75C1            0.75C1      0.75C1

Now C1-0.75C1=1 ===>C1=4==>[hcoo-]=3

Now  HCOOH----><----HCOO-   +   H+

        0.2-x                 x+y            x

y=added HCOO-

when y=0 x=6.4*10^(-3)

==>when y=3 x is further suppressed and x can be neglected in comparison with y as well as 0.2.

So we have  x*y/0.2=Ka=2.4*10^(-4)

yielding x=1.6*10^(-5)=[H+]