JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 20)
Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10$$-$$5, the ratio of salt to acid concentration in the buffer solution will be :
4 : 5
1 : 10
10 : 1
5 : 4
Explanation
HA $$\rightleftharpoons$$ H+ + A$$-$$
Ka = $${{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$$ = 10$$-$$5
pH = pKa + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$
$$ \Rightarrow $$$$\,\,\,$$ 6 = $$-$$ log [10$$-$$5] + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$
$$ \Rightarrow $$$$\,\,\,$$ 6 = 5 + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$
$$ \Rightarrow $$ log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 1
$$ \Rightarrow $$$$\,\,\,$$ $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 10
$$\therefore\,\,\,$$ Salt : Acid = 10 : 1
Ka = $${{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$$ = 10$$-$$5
pH = pKa + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$
$$ \Rightarrow $$$$\,\,\,$$ 6 = $$-$$ log [10$$-$$5] + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$
$$ \Rightarrow $$$$\,\,\,$$ 6 = 5 + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$
$$ \Rightarrow $$ log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 1
$$ \Rightarrow $$$$\,\,\,$$ $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 10
$$\therefore\,\,\,$$ Salt : Acid = 10 : 1
Comments (0)
