PLEASE HELP ASAP!

Question

calculate the PH of the solution when 0.5 g of NaOH is added to 200 ml of 0.1 M HCOOH(formic acid). It will be accepted that there is no change in the volume of solution when solid substance is added (Ka=1.8*10^(-4) ), (Na=22.99 g/mol, O=16 g/mol, H=1.01 g/mol)

PLEASE WRITE THE SOLUTION WITH ALL STEPS

Answer 1

Write the reaction that is taking place:

NaOH + HCOOH ==> HCOONa + H2O (acid + base neutralization) ... balanced equation

Next, find out how much of each reactant is present:

moles NaOH = 0.5 g NaOH x 1 mol NaOH/ 40 g = 0.0125 moles NaOH

moles HCOOH = 200 ml x 1 L/1000 ml x 0.1 mol/L = 0.020 moles HCOOH

Since they react in a 1:1 mole ration (see balanced equation), there will be some HCOOH left over, How much?

0.020 moles - 0.0125 moles = 0.0075 moles HCOOH left over.

Since the volume is the same, we can find the molar concentration of HCOOH as

0.0075 moles/0.200 L = 0.0375 M HCOOH

Using the Ka for this weak acid, we can now find the [H⁺] and then the pH.

Ka = [H⁺][COO^-] / [HCOOH]

1.8x10^-4 = (x)(x) / (0.0375 -x) and if we assume x is small compared to 0.0375, we can ignore it

1.8x10^-4 = x²/0.0375

x² = 6.75x10^-6

x = 2.6x10^-3 M = [H⁺] (note: this value is about 7% of 0.0375 so we should go back and use quadratic)

pH = -log [H⁺] = -log 2.6x10^≈-3

pH ≈ 2.6

(I'll leave it to you to go back and use the quadratic to solve for x and take the negative log of that value(