Question

Calculate the pH of a solution formed by mixing 65 mL of 0.34 M NaHCO3 with 75 mL of 0.10 M Na2CO3.

Answer 1

we know that

moles = concentration x volume (ml) / 1000

so

moles of NaHC03 = 0.34 x 65 / 1000 = 0.0221

moles of Na2C03 = 0.1 x 75 / 1000 = 0.0075

now

we NaHC03 and Na2C03 are acid / conjugate base system

so

they form a buffer

and for buffers

according to henderson hasselbalch equation

pH = pKa + log [conjugate base / acid]

in this case

pH = pKa + log [Na2C03 / NaHC03]

pKa for NaHC03 is 10.32

so

pH = 10.32 + log [Na2C03 / NaHC03]

we know that

molarity = moles / volume (L)

as the final volume is same for both

ratio of molarity = ratio of moles

so

pH = 10.32 + log [0.0075 /0.0221]

pH = 9.85067

so

pH of the given solution is 9.85067