Question

How many grams of solid $NaOH$ must be added to $100ml$ of a buffer solution which is $0.1M$ each with respect to Acid $HA$ and salt $N{a}^{+}{A}^{-}$ to make the $pH$ of the solution $5.5$? ( Given $p{k}_a\left(HA\right)=5$)

• A. $2.08\times {10}^{-1}$
• B. $2.10\times {10}^{-1}$
• C. $2.01\times {10}^{-2}$
• D. None of these

Answer 1

Answer: (A)

$2.08\times {10}^{-1}$

$pH=p{k}_a+log\frac{\left[A^{-}\right]}{\left[HA\right]}$

$5.5=5+log\frac{\left[A^{-}\right]}{\left[HA\right]}$

$log\frac{\left[A^{-}\right]}{\left[HA\right]}=0.5$

$\frac{\left[A^{-}\right]}{\left[HA\right]}=3.16$ ....(1)

Let $W$ moles of $NaOH$ are added

$\left[NaOH\right]=\frac{wmol}{100ml}\times \frac{1000ml}{1L}=10wM$

$NaOH+HA=NaOH+H_{2}O$

$10w$ molar $NaOH$ will combine with

$10w$ molar $HA$ to form $10wMA$

$\left[A^{-}\right]=(0.1+10w)M$

$\left[HA\right]=(0.1-10w)M$

$\frac{\left[A^{-}\right]}{\left[HA\right]}=\frac{0.1+10w}{0.1-10w}$ ...(2)

$\left(1\right)=\left(2\right)$

$\frac{0.1+10w}{0.1-10w}=3.16$

$0.1+10w=3.16(0.1-10w)$

$0.1+10w=0.316-316w$

$71.6w=0.216$

$w=0.00519$ mole

Molecular weight of $NaOH=40g/mol$, mass of $NaOH=0.00519\times 40=2.08\times {10}^{-1}g$

Ans. is option (A).