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ChemMedMain 2026 • 4th Apr - Shift 1

Calculate pH of Ammonium Hydroxide and Ammonium Chloride Buffer Solution

The pH\text{pH} of a solution obtained by mixing 5 mL5\text{ mL} of 0.1 M NH4OH0.1\text{ M } \text{NH}_4\text{OH} solution with 250 mL250\text{ mL} of 0.1 M NH4Cl0.1\text{ M } \text{NH}_4\text{Cl} solution is \underline{\quad} ×10−2\times 10^{-2}. (Nearest integer) Given: pKb(NH4OH)=4.74\text{pK}_b (\text{NH}_4\text{OH}) = 4.74

log⁡2=0.30\log 2 = 0.30
log⁡3=0.48\log 3 = 0.48
log⁡5=0.70\log 5 = 0.70

Official Numerical Answer756

Topics & Concepts

EquilibriumAcid-Base Equilibrium

Step-by-Step Solution

To determine the pH\text{pH} of the buffer solution formed by mixing ammonium hydroxide (NH4OH\text{NH}_4\text{OH}) and ammonium chloride (NH4Cl\text{NH}_4\text{Cl}), we use the Henderson-Hasselbalch equation for a basic buffer.

Step 1: Calculate the millimoles of the weak base and its conjugate acid (salt)

  • Millimoles of weak base (NH4OH\text{NH}_4\text{OH}): moles of NH4OH=Volume (in mL)×Molarity\text{moles of } \text{NH}_4\text{OH} = \text{Volume (in mL)} \times \text{Molarity} Millimoles of NH4OH=5 mL×0.1 M=0.5 mmol\text{Millimoles of } \text{NH}_4\text{OH} = 5\text{ mL} \times 0.1\text{ M} = 0.5\text{ mmol}

  • Millimoles of salt (NH4Cl\text{NH}_4\text{Cl}): moles of NH4Cl=Volume (in mL)×Molarity\text{moles of } \text{NH}_4\text{Cl} = \text{Volume (in mL)} \times \text{Molarity} Millimoles of NH4Cl=250 mL×0.1 M=25 mmol\text{Millimoles of } \text{NH}_4\text{Cl} = 250\text{ mL} \times 0.1\text{ M} = 25\text{ mmol}

Step 2: Calculate the pOH\text{pOH} of the solution

The Henderson-Hasselbalch equation for a basic buffer is: pOH=pKb+log⁡10([Salt][Base])\text{pOH} = \text{pK}_b + \log_{10} \left( \frac{[\text{Salt}]}{[\text{Base}]} \right)

Since both the salt and the base are in the same total volume, the ratio of concentrations equals the ratio of millimoles: pOH=pKb+log⁡10(Millimoles of NH4ClMillimoles of NH4OH)\text{pOH} = \text{pK}_b + \log_{10} \left( \frac{\text{Millimoles of } \text{NH}_4\text{Cl}}{\text{Millimoles of } \text{NH}_4\text{OH}} \right)

Substitute the given values into the equation: pOH=4.74+log⁡10(250.5)\text{pOH} = 4.74 + \log_{10} \left( \frac{25}{0.5} \right) pOH=4.74+log⁡10(50)\text{pOH} = 4.74 + \log_{10}(50)

Using the logarithmic identity log⁡10(50)=log⁡10(10×5)=log⁡10(10)+log⁡10(5)\log_{10}(50) = \log_{10}(10 \times 5) = \log_{10}(10) + \log_{10}(5): log⁡10(50)=1+0.70=1.70\log_{10}(50) = 1 + 0.70 = 1.70

Now, calculate pOH\text{pOH}: pOH=4.74+1.70=6.44\text{pOH} = 4.74 + 1.70 = 6.44

Step 3: Calculate the pH\text{pH} of the solution

At 25∘C25^\circ\text{C}, the relationship between pH\text{pH} and pOH\text{pOH} is: pH+pOH=14\text{pH} + \text{pOH} = 14 pH=14−6.44=7.56\text{pH} = 14 - 6.44 = 7.56

Step 4: Express the answer in the required format

The question asks for the pH\text{pH} value in terms of ‾×10−2\underline{\quad} \times 10^{-2}: pH=7.56=756×10−2\text{pH} = 7.56 = 756 \times 10^{-2}

Thus, the required integer value is 756756.

Calculate pH of Ammonium Hydroxide and Ammonium Chloride Buffer Solution | Chemistry PYQ Solution - JEE Challenger