Question

Question 1.


A. With the help of equation show that the aqueous solution of:


i. Ammonia is basic


ii. Hydrogen carbonate is acidic



B. Calculate the pH of the following;


i. 0.01molL⁻1of hydrochloric acid (HCl)


ii. 0.02molL⁻1 of methanoic (HCOOH) acid given Ka (HCOOH) = 1.78 x 10⁻4



C. A pH value can be calculated from [OH⁻] and Kw


i. Calculate the concentration of OH⁻ (aq) in a solution of HCl with a pH of 3.7, given Kw = 1 x 10⁻14


ii. Give formula for the conjugate base of HCl

Answer 1

Solution.

A.

i. $NH_3+H_2O\to NH_4OH;$

ii. $H_2CO_3+2H_2O\to2H_3O+CO_3;$

B.

i. $[HCl]=0.01molL^{-1};$

$pH=-log[H_3O^+];$

$pH=-log0.01=2;$

ii.$[HCOOH]=0.02molL^{-1};$

$K_\alpha(HCOOH)=1.78\sdot10^{-4};$

$K_\alpha=\dfrac{[H_3O^+][HCO_2^-]}{[HCO_2H]};$

$[H_3O^+]=[HCO_2^-]=x;$

$x^2=K_\alpha[HCO_2H];$

$x^2=1.78\sdot10^{-4}\sdot0.02=0.0356\sdot10^{-4}; x=0.189\sdot10^{-2}$ ;

$[H_3O^+]=0.189\sdot10^{-2};$

$pH=-log(0.189\sdot10^{-2})=log529=2.72;$

C.

i. $K_b=1\sdot10^{-4};$

$pH=3.7;$

$pH=-log[H_3O];\implies[H_3O^+]=10^{-pH};$

$[H_3O^+]=10^{-3.7}=2\sdot10^{-4};$

$K_b=\dfrac{[OH^-][HCl]}{[Cl^-]};$

$[OH^-]=[HCl]=x;$

$x^2=K_b[Cl^-];$

$x^2=1\sdot10^{-4}\sdot 2\sdot10^{-4}=2\sdot10^{-4}; x=1.41\sdot10^{-2};$

$[OH^-]=1.41\sdot10^{-2};$

ii. formula for the conjugate base of $HCl$ : $Cl^-;$

Answer: A. i.$NH_3+H_2O\to NH_4OH;$

ii.$H_2CO_3+2H_2O\to2H_3O+CO_3;$

B. i. $pH=2;$

ii. $pH=2.72;$

C. i. $[OH^-]=1.41\sdot10^{-2};$

ii. formula for the conjugate base of $HCl$ is $Cl^-.$