Question

Calculate the ph of a buffer solution prepared by dissolving 30g of na2co3 in 500 ml of an aqueous solution containing 150 ml of 1m hcl. ka for hco-3 = 5.63 x 10-11

Answer 1

what is ph of buffer solution??????

Answer 2

Answer: pH of buffer solution is 10.19

Explanation:

$Na_2CO_3+HCl\rightarrow NaCl+NaHCO_3$

Moles of $Na_2CO_3$ = $\frac{\text{Given mass}}{\text{Molar mass}}= \frac{30g}{106g/mole}=0.28moles$

$Moles of HCl=Molarity\times {\text {Volume in L}}=1M\times 0.15L=0.15moles$

Thus As 1 mole of $HCl$ combines with 1 mole of $Na_2CO_3$

0.15 moles of $HCl$ will combines with 0.15 moles of $Na_2CO_3$

Thus $HCl$ is the limiting reagent as it limits the formation of products and $Na_2CO_3$ is an excess reagent.

And 0.15 moles of $HCl$ will produce 0.15 moles of $NaHCO_3$

Moles of $Na_2CO_3$ left= (0.23-0.15) = 0.08 moles

The total volume of the solution will be

$V_{total}=500mL+150mL=650mL=0.65L$

According to the Henderson-Hasselbach equation, pH of such buffer solution can be calculated as:

$pH=p_{K_a}+log\frac{[CO_3^{2-}]}{[HCO_3^{-}]}$

${[CO_3^{2-}]}=\frac{moles}{\text {Volume in L}}=\frac{0.08}{0.65L}=0.12M$

${[HCO_3^-]}=\frac{moles}{\text {Volume in L}}=\frac{0.15}{0.65L}=0.23M$

$p_{K_a}=-log[k_a]$

Thus $p_{K_a}=-log([5.63\times 10^{-11}])=10.24$

$pH=10.24+log\frac{[0.08]}{[0.15]}$

$pH=10.19$