20 mL of 0.1 MH2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resulatant mixture is: [pkb of NH4OH = 4.7].

Question

20 mL of 0.1 MH2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resulatant mixture is: [pkb of NH4OH = 4.7]. (A) 9.4 (B) 5.0 (C) 9.0 (D) 5.2

Tardigrade Admin · Accepted Answer

20 ml 0 .1 M H 2 SO 4 ⇒ η H +=4 30 ml 0 .2 M NH 4 OH ⇒ η NH 4 OH =6 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/525bb1b576f24670019850dbeb083676-.png" /> pOH =pKb + log (NH+4/NH4OH) = 4.7 + log 2 = 4.7 + 0.3 = 5 pH = 14 - 5 = 9

Q. $20 m L$ of $0.1 M H_{2} S O_{4}$ solution is added to $30 m L$ of $0.2 M N H 4 O H$ solution. The $p H$ of the resulatant mixture is : $[p kb o f N H_{4} O H = 4.7]$ .

9.4

5.0

9.0

5.2

$20 m l 0.1 M H_{2} S O_{4} \Rightarrow η_{H^{+}} = 4$
$30 m l 0.2 M N H_{4} O H \Rightarrow η_{N H_{4} O H} = 6$

$pO H = p K_{b} + lo g \frac{N H _{4}^{+}}{N H _{4} O H}$
$= 4.7 + lo g 2$
$= 4.7 + 0.3 = 5$
$p H = 14 - 5 = 9$

Q. 20mL of 0.1MH2​SO4​ solution is added to 30mL of 0.2MNH4OH solution. The pH of the resulatant mixture is : [pkbofNH4​OH=4.7].

9.4

5.0

9.0

5.2

20ml0.1MH2​SO4​⇒ηH+​=4 30ml0.2MNH4​OH⇒ηNH4​OH​=6 pOH=pKb​+logNH4​OHNH4+​​ =4.7+log2 =4.7+0.3=5 pH=14−5=9

Q. $20 m L$ of $0.1 M H_{2} S O_{4}$ solution is added to $30 m L$ of $0.2 M N H 4 O H$ solution. The $p H$ of the resulatant mixture is : $[p kb o f N H_{4} O H = 4.7]$ .

$20 m l 0.1 M H_{2} S O_{4} \Rightarrow η_{H^{+}} = 4$
$30 m l 0.2 M N H_{4} O H \Rightarrow η_{N H_{4} O H} = 6$

$pO H = p K_{b} + lo g \frac{N H _{4}^{+}}{N H _{4} O H}$
$= 4.7 + lo g 2$
$= 4.7 + 0.3 = 5$
$p H = 14 - 5 = 9$