Question

Calculate the pH of a buffer solution prepared by dissolving 4.2 g of NaHCO3 and 5.3 g of Na2CO3 in 0.20 L of water. Will the pH change if the solution volume is increased by a factor of 10?

Answer 1

Solution :-

Given

Mass of NaHCO3 = 4.2 g

Mass of Na2CO3 = 5.3 g

Volume = 0.20 L

Lets first calculate the moles of the NaHCO3 and Na2CO3

Formula to calculate moles is

Moles = mass in gram / molar mass

Moles of NaHCO3 = 4.2 g / 84.007 g per mole

                                = 0.05 mol

Moles of Na2CO3 = 5.3 g /105.99 g per mol

                              = 0.05 mol

Now lets calculate molarity of the both NaHCO3and Na2CO3

Molarity = moles / liter

Molarity of NaHCO3 = 0.05 mol / 0.20 L = 0.25 M

Molarity of Na2CO3 = 0.05 mol / 0.20 L = 0.25 M

HCO₃^- is the acid and CO₃^2- is the conjugate base

Now using the Henderson Hasel Balch equation we can calculate the pH

pH= pka + log ([base]/[acid])

so here we need the pka of the HCO₃^-

using the ka of HCO₃^- we can find the pka

ka of HCO₃^- = 4.7*10^-11

pka = - log [ ka ]

pka = - log [4.7*10^-11]

pka = 10.3

now lets put the values in the Henderson equation

pH = 10.3 + log ([0.25]/[0.25])

pH = 10.3 + 0

pH= 10.3

pH of the solution will not change even if the volume increased by factor 10

because the ratio of the base to acid is 1

therefore pH = pKa