Question

Calculate the pH, pOH, [OH-] and [H30+] for 2.75 x 10^-2 M Ba(OH)2(aq) solution.

You have 200.0mL of a buffer solution containing 0.175 M HCO2H and 0.225 M NaHCO2. What is the pH after 25.0mL of 0.300 M KOH is added to this buffer solution? For HCO2H, Ka= 1.8 x 10^-4.

Answer 1

molarity of Ba(OH)₂ = 2.75X10^-2

molarity = number of moles of solute / volume of solution

As Ba(OH)₂ is base we get pOH

pOH = - log [OH^-] pOH = - log 0.0275 =1.56

here [OH^-] = 0.0275 moles

we know that pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.56 = 12.44

now pH = - log [H₃O⁺]

12.44 = - log [H₃O⁺]

log [H₃O⁺] = -12.44

[H₃O⁺] = 3.6X10^-13 M

volume of buffer = 200 mL = 0.2 L

molarity of HCOOH = 0.175

molarity of HCOONa= 0.225

pH = pKa + log [[A-]/ [HA]]

given for HCOOH, Ka = 1.8x10^-4

pKa = - log Ka = - log 1.8x10^-4 = 3.744

normally formic acid in aqueous solution undergo reaction like as below

HCOOH +H₂O ---> H₃O⁺ + HCOO^-

Ka = 1.8x10^{-4 ,} [H₃O⁺] = 1.43X10^-4

pH = - log [H₃O⁺] pH = - log 1.43X10^-4 = 3.84

now 25ml of 0.300 M KOH is added as KOH is a strong base it will be completely neutralised by the acid. the reaction is give as below

HCOOH + OH^- H₂O +HCOO^-

Concentration of acid is 0.175M X 200 mL =35mmol

concentration of base is 0.300MX 25mL =7.5mmol

Concentration of mixture reduces the concentration of acid given by 35 - 7.5 = 27.5mmol

Concentration of conjugate base HCOO^- is given by (0.225M X200) + 7.5 = 45 + 7.5 = 52.5mmol

this forms a buffer solution now the pH is given by

pH = p Ka + log [HCOO^-]/ [HCOOH]

pKa = - log Ka = - log 1.84X10^-4 = 3.735

therefore pH = 3.735 + log (52.5/27.5) = 4.016

Hence initial pH is 3.84 by adding strong base the pH is increased to 4.016.