The pH of a solution obtained by mixing 5 mL of 0.1 M NH4OH solution with 250 mL of 0.1 M NH4Cl solution is _______ x 10^-2. (Nearest integer) Given: pKb (NH4OH) = 4.74 log 2 = 0.30 log 3 = 0.48 log 5 = 0.70

The correct answer is Option 756. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q74 – JEE-MAIN 2026

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Home Chemistry EquilibriumQuestion #421

The pH of a solution obtained by mixing

5 mL

0.1 {M NH}_{4} OH

solution with

250 mL

0.1 {M NH}_{4} Cl

solution is _______

\times 10^{- 2}

. (Nearest integer) Given:

p K_{b} ({NH}_{4} OH) = 4.74

\log 2 = 0.30

\log 3 = 0.48

\log 5 = 0.70

Enter Calculated Numerical Value:

Clue 1 of 3: Identify the type of solution

This solution contains a weak base ( $N H_{4} O H$ ) and its conjugate acid ( $N H_{4}^{+}$ from $N H_{4} C l$ ), which forms a basic buffer.

🧠 Full Solution Path

Step 1: Identify Buffer Components and Calculate Moles✦ Active

The solution is a basic buffer formed by $N H_{4} O H$ (weak base) and $N H_{4} C l$ (source of conjugate acid $N H_{4}^{+}$ ). First, calculate the moles of each component:

Moles of N H_{4} O H = 5 mL \times 0.1 M = 0.5 mmol

Moles of N H_{4}^{+} = Moles of N H_{4} C l = 250 mL \times 0.1 M = 25 mmol

The total volume of the solution after mixing is $5 mL + 250 mL = 255 mL$ . The concentrations are then $[N H_{4} O H] = \frac{0.5}{255} M$ and $[N H_{4}^{+}] = \frac{25}{255} M$ .

Step 2: Calculate pOH using Henderson-Hasselbalch Equation○ Expand

Given $p K_{b} = 4.74$ . Use the Henderson-Hasselbalch equation for a basic buffer:

p O H = p K_{b} + \log \frac{[conjugate acid]}{[weak base]}

p O H = 4.74 + \log \frac{[N H_{4}^{+}]}{[N H_{4} O H]}

p O H = 4.74 + \log \frac{25 / 255}{0.5 / 255}

p O H = 4.74 + \log \frac{25}{0.5}

p O H = 4.74 + \log 50

Using the given $\log$ values ( $\log 5 = 0.70$ ):

\log 50 = \log (5 \times 10) = \log 5 + \log 10 = 0.70 + 1 = 1.70

Substitute this value back into the pOH equation:

p O H = 4.74 + 1.70 = 6.44

💡 Teacher's Secret Hint

Remember that the volume terms cancel out in the log ratio when calculating pOH for a buffer.

Step 3: Calculate pH and Determine Final Answer○ Expand

At $25^{\circ} C$ , the relationship between pH and pOH is $p H + p O H = 14$ . Calculate the pH:

p H = 14 - p O H = 14 - 6.44 = 7.56

The question asks for the pH in the format _______ $\times 10^{- 2}$ (Nearest integer). Let $p H = X \times 10^{- 2}$ :

7.56 = X \times 10^{- 2}

X = 7.56 \times 100 = 756

The nearest integer is 756.

💡 Teacher's Secret Hint

Pay attention to the requested format for the final answer, including the power of 10 and rounding to the nearest integer.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 74 – JEE-MAIN 2026

Clue 1 of 3: Identify the type of solution

🧠 Full Solution Path