Question

1.0 L of a 1.2 M solution of formic acid (Ka=1.8x10-4) was prepared. What is the pH of this solution? 1.8328 You are correct. Your receipt no. is 162-1608 Previous Tries Enough potassium formate was added to achieve a final potassium formate concentration of 1.7 M (assume the volume does not change). What is the pH of this new solution? 3.8960 You are correct. ceipt no. is 162-6598 Previous Tries 10 mL of 10 M NaOH is added to the solution. What is the new pH? 3.783 Submit Answer Incorrect. Tries 1/2 Previous Tries

Answer 1

From 2nd part of the question we got [H+] =1.27x10-4 M

In 1L solution,

number of moles of H+ ions = molarity x volume of solution (Molarity= no. of moles/volume of solution in L)

=1.27 x 10-4 x 1

= 1.27 x 10-4  moles

Acoording to question,

10 ml of 10M NaOH solution is added to the solution which will neautralise equal amount of H+ ions.

Number of moles of OH- ions = molarity x volume( in L)

= 10M x 10/1000

= 0.1 moles

As number of moles of OH- ions are greater than H+ ions so excess of OH- ions will be present in solution and the resulting solution will be basic in nature.

Excess of moles of OH- ions present are = 0.1 - (1.27 x 10-4)

= 0.0998 moles

Total volume = 1L + 0.01L(10ml)

= 1.01L

Molar concentration of OH- ions left = no. of moles/volume of solution

= 0.0998/1.01

= 0.0988 M

Now lets calculate pOH first and then with the help of that we will calculate pH

pOH = - log[OH-]

= - log[0.0988]

= 1.005

We know that,

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.005

pH = 12.995