Question

The amount of calcium formate which should be added in 500ml of 0.1M HCOOH to obtain a solution with pH=4 is [given $K_{a}ofHCOOH=1.8\times {10}^{-4}$] :

• A. 4.36g
• B. 5.85g
• C. 3.76g
• D. 5.12g

Answer 1

The correct option is B 5.85g

Its a buffer of $\left[HCOOH\right]$ and $\left[HCOO\right]$

acid formate sale of calcium

Heudesson- Hasselbalch Equation for Acid Buffer

$pH=p{k}_A+\log \left[\frac{\left[salt\right]}{\left[acid\right]}\right]$

$4=(4-\log 1.8)+\log \frac{\left[HCO{O}^{-}\right]}{0.1}$

$\Rightarrow \left[HCO{O}^{-}\right]=0.18mol{l}^{-1}$

$\Rightarrow \because v=0.5l$ and $Ca{\left[HCOO\right]}_2$ is the salt

$\Rightarrow$ Amount of salt $=0.18\times 0.5\times \frac{1}{2}$

$=0.045$

$0.045mol$

$0.045\times 130$ (molar mass)

$=5.85g$ (Calcium formate $=130gm{r}^{-1}$)