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Question: How many gram of solid \(NaOH\) must be added to 100 ml of a buffer solution which is 0.1M each w.r....

How many gram of solid NaOHNaOH must be added to 100 ml of a buffer solution which is 0.1M each w.r.t acid and salt Na+A−N{{a}^{+}}{{A}^{-}} to make pH of the solution 5.5. Given pKa(HA)=5p{{K}_{a}}(HA)=5 (use antilog (0.5)=3.16)
(A)2.08×1012.08\times {{10}^{1}}
(B)3.05×1033.05\times {{10}^{3}}
(C)2.01×1022.01\times {{10}^{2}}
(D) None of these

Explanation

Solution

Hint Buffer solution is a solution that resists the change in pH and pH does not change significantly in addition to a small amount of acid and base. pH of a buffer solution is calculated by Henderson’s equation pH=pKa+log⁡[salt][acid]pH=p{{K}_{a}}+\log \dfrac{[salt]}{[acid]}. If a small quantity of base is added to the buffer solution, the OH−O{{H}^{-}} ion are cancelled by the acid component HA+NaOHA−Na++H2OHA+NaOH{{A}^{-}}N{{a}^{+}}+{{H}_{2}}O

Complete Step by step solution:
An acid buffer solution is the mixture of weak acid and salt of the same weak acid with any strong base so, in this given question weak acid is and its salt is. So they dissociate in the following manner in the buffer solution –
HAH++A−HA{{H}^{+}}+{{A}^{-}}
Na+ANa++A−N{{a}^{+}}AN{{a}^{+}}+{{A}^{-}}
To solve this question first we will calculate the number of mole of weak acid HA in 0.1M solution,
Since,M=nV(L)M=\dfrac{n}{V(L)}
For 0.1M HA,0.1=nHA100ml×10000.1=\dfrac{{{n}_{HA}}}{100ml}\times 1000
⇒nHA=0.01mole\Rightarrow {{n}_{HA}}=0.01mole
After addition of NaOHNaOH suppose xx mole of NaOHNaOH is added to the 100 ml buffer solution containing 0.1 mole of HA and Na+A−N{{a}^{+}}{{A}^{-}}. So, xx mole of NaOHNaOH will neutralise thexxmole of HA. Neutralization will increase the concentration ofNa+A−N{{a}^{+}}{{A}^{-}}and decrease the concentration of HA in the following manner in the buffer solution.
[HA]=0.01−x100ml[HA]=\dfrac{0.01-x}{100ml}, [Na+A−]=0.01+x100ml[N{{a}^{+}}{{A}^{-}}]=\dfrac{0.01+x}{100ml}
So after applying the Henderson’s equation we will calculate the value ofxx
pH=pKa+log⁡[salt][acid]pH=p{{K}_{a}}+\log \dfrac{[salt]}{[acid]}
⇒5.5=5+[0.01+x][0.01−x]\Rightarrow 5.5=5+\dfrac{[0.01+x]}{[0.01-x]}
This, will be equal tolog⁡[0.01+x][0.01−x]=0.5\log \dfrac{[0.01+x]}{[0.01-x]}=0.5
By, taking base of 10 on both the sides, we get
log⁡10[0.01+x][0.01−x]=100.5{{\log }_{10}}\dfrac{[0.01+x]}{[0.01-x]}={{10}^{0.5}}
⇒0.01+x0.01−x=3.16\Rightarrow \dfrac{0.01+x}{0.01-x}=3.16 [ Since, antilog(0.5) = 3.16]
Thus, we get the value of xx as,
x=0.0316−0.014.16=0.0052x=\dfrac{0.0316-0.01}{4.16}=0.0052
Here xx represents the moles of sodium hydroxide, so we will calculate the weight of NaOHNaOH by applying following mole formula that is,
Number of mole (n) =weightmol.wt.\dfrac{weight}{mol.wt.}
Weight of NaOHNaOH= n×\times molecular weight
Thus, by substituting the values,
Weight of NaOHNaOH=0.052×40=0.207g0.052\times 40=0.207g
The above value can also be written as,2.07×10−2g2.07\times {{10}^{-2}}g

So, option (D) will be the correct answer.

Note: The concentration of H+{{H}^{+}} ions in the buffer solution is high, hence most of the added OH−O{{H}^{-}} ions are consumed and thus the addition of strong base do not causes appreciable change in the pH of the buffer solution. In neutralization chemical reactions equal mole of acid and base neutralize each other.