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Ionic Equilibrium question

2021 · 26 Feb · Shift 2 · Q22
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  5. /2021 · 26 Feb · Shift 2 · Q22

Ionic Equilibrium question

2021 · 26 Feb · Shift 2 · Q22

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The pH of ammonium phosphate solution, if pka of phosphoric acid and pkb of ammonium hydroxide are 5.23 and 4.75 respectively, is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 7

  1. Identify the salt and ions present

Ammonium phosphate here is formed from:

  • weak base: ammonium hydroxide, NH4OH\mathrm{NH_4OH}NH4​OH
  • weak acid: phosphoric acid, taking the relevant acidic step with given pKa=5.23pK_a = 5.23pKa​=5.23

For a salt of a weak acid and weak base, the pH is given by:

pH=7+12log⁡(KbKa)\mathrm{pH} = 7 + \frac{1}{2}\log\left(\frac{K_b}{K_a}\right)pH=7+21​log(Ka​Kb​​)

Using pKpKpK values, this becomes:

pH=7+12(pKa−pKb)\mathrm{pH} = 7 + \frac{1}{2}(pK_a - pK_b)pH=7+21​(pKa​−pKb​)

because

log⁡(KbKa)=−pKb−(−pKa)=pKa−pKb\log\left(\frac{K_b}{K_a}\right)= -pK_b -(-pK_a)=pK_a-pK_blog(Ka​Kb​​)=−pKb​−(−pKa​)=pKa​−pKb​
  1. Substitute the given values

Given:

pKa=5.23,pKb=4.75pK_a = 5.23, \qquad pK_b = 4.75pKa​=5.23,pKb​=4.75

So,

pH=7+12(5.23−4.75)\mathrm{pH} = 7 + \frac{1}{2}(5.23 - 4.75)pH=7+21​(5.23−4.75) pH=7+12(0.48)\mathrm{pH} = 7 + \frac{1}{2}(0.48)pH=7+21​(0.48) pH=7+0.24=7.24\mathrm{pH} = 7 + 0.24 = 7.24pH=7+0.24=7.24
  1. Integer answer

Since this is an integer-type question, the pH is approximately:

7\boxed{7}7​
  1. Compare with stored answer

Stored correct answer: 777

Our derived value is 7.247.247.24, which rounds to the integer 777. Hence it agrees with the stored answer.

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