A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of an N / 20 HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as indicator required 25 mL of the same acid. the amount of KOH present in the solution is

Question

A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of an N / 20 HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as indicator required 25 mL of the same acid. the amount of KOH present in the solution is (A) 0.014 g (B) 0.14 g (C) 0.028 g (D) 1.4 g

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c754e3f971dbcd13fd45320609868238-.png" /> i. (x × 1)+(y × 2) × (1/2)=(1/20) × 15 ii. (x × 1)+(y × 2)=(1/20) × 25 ⇒ y=0.5 and x= 0.25 ⇒ KOH =x mmoles =0.25 × 10-3 × 56 g = 0.014 g

$0.014 g$

$0.14 g$

$0.028 g$

$1.4 g$

i. $(x \times 1) + (y \times 2) \times \frac{1}{2} = \frac{1}{20} \times 15$
ii. $(x \times 1) + (y \times 2) = \frac{1}{20} \times 25$
$\Rightarrow y = 0.5$ and $x =$ $0.25$
$\Rightarrow K O H = x$ mmoles $= 0.25 \times 1 0^{- 3} \times 56 g =$ $0.014 g$

0.014g

0.14g

0.028g

1.4g

i. (x×1)+(y×2)×21​=201​×15 ii. (x×1)+(y×2)=201​×25 ⇒y=0.5 and x= 0.25 ⇒KOH=x mmoles =0.25×10−3×56g= 0.014g

$0.014 g$

$0.14 g$

$0.028 g$

$1.4 g$

i. $(x \times 1) + (y \times 2) \times \frac{1}{2} = \frac{1}{20} \times 15$
ii. $(x \times 1) + (y \times 2) = \frac{1}{20} \times 25$
$\Rightarrow y = 0.5$ and $x =$ $0.25$
$\Rightarrow K O H = x$ mmoles $= 0.25 \times 1 0^{- 3} \times 56 g =$ $0.014 g$