The pH of a solution obtained by mixing 250 mL of 0.5 M NaOH, 100 mL of 0.5 M HCl and 400 mL of water is

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q131 – AP-EAMCET 2026

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Home Chemistry EquilibriumQuestion #413

The pH of a solution obtained by mixing

250 mL

0.5 M NaOH

100 mL

0.5 M HCl

and

400 mL

of water is

Clue 1 of 3: Acid-Base Neutralization

Understand that strong acids and strong bases react completely to form water. Determine the initial moles of acid and base present.

🧠 Full Solution Path

Step 1: Calculate initial moles of acid and base✦ Active

First, calculate the initial moles of ${OH}^{-}$ from $NaOH$ and $H^{+}$ from $HCl$ . For $NaOH$ : Volume ( $V_{NaOH}$ ) = $250 mL = 0.250 L$ Molarity ( $M_{NaOH}$ ) = $0.5 M$

Moles of NaOH = M_{NaOH} \times V_{NaOH} = 0.5 M \times 0.250 L = 0.125 mol

Since $NaOH$ is a strong base, it completely dissociates, so moles of ${OH}^{-} = 0.125 mol$ . For $HCl$ : Volume ( $V_{HCl}$ ) = $100 mL = 0.100 L$ Molarity ( $M_{HCl}$ ) = $0.5 M$

Moles of HCl = M_{HCl} \times V_{HCl} = 0.5 M \times 0.100 L = 0.050 mol

Since $HCl$ is a strong acid, it completely dissociates, so moles of $H^{+} = 0.050 mol$ .

💡 Teacher's Secret Hint

Remember to convert volumes from milliliters to liters before calculating moles to ensure consistent units.

Step 2: Determine the moles of excess reactant after neutralization○ Expand

The neutralization reaction between a strong acid and a strong base is: $H^{+} (a q) + {OH}^{-} (a q) \to H_{2} O (l)$ This reaction occurs in a $1 : 1$ molar ratio. Compare the moles of $H^{+}$ and ${OH}^{-}$ to find the limiting reactant and the excess moles.

{Moles of OH}^{-} available = 0.125 mol

{Moles of H}^{+} available = 0.050 mol

Since the moles of $H^{+}$ are less than the moles of ${OH}^{-}$ , $H^{+}$ is the limiting reactant. It will completely react with an equal amount of ${OH}^{-}$ . The moles of ${OH}^{-}$ remaining in the solution are:

{Moles of OH}^{-} remaining = {Initial moles of OH}^{-} - {Moles of H}^{+} reacted

{Moles of OH}^{-} remaining = 0.125 mol - 0.050 mol = 0.075 mol

💡 Teacher's Secret Hint

Always identify the limiting reactant in neutralization problems to correctly determine which species remains in excess.

Step 3: Calculate the total volume of the solution○ Expand

Sum the volumes of all liquids mixed to find the final total volume of the solution.

Total Volume = V_{NaOH} + V_{HCl} + V_{water added}

Total Volume = 250 mL + 100 mL + 400 mL = 750 mL

Convert the total volume to liters:

Total Volume = 750 mL = 0.750 L

💡 Teacher's Secret Hint

Don't forget to include any added water in the total volume calculation, as it dilutes the final concentration.

Step 4: Calculate the concentration of the excess ion○ Expand

Now, calculate the molar concentration of the remaining ${OH}^{-}$ ions in the final solution.

[{OH}^{-}] = \frac{{Moles of OH}^{-} remaining}{Total Volume (in L)}

[{OH}^{-}] = \frac{0.075 mol}{0.750 L} = 0.1 M

Step 5: Calculate pOH and then pH○ Expand

Since we have the concentration of ${OH}^{-}$ , we can calculate the pOH first.

pOH = - \log_{10} [{OH}^{-}]

pOH = - \log_{10} (0.1) = - \log_{10} (10^{- 1}) = 1

Finally, use the relationship between pH and pOH at $25^{\circ} C$ to find the pH.

pH + pOH = 14

pH = 14 - pOH = 14 - 1 = 13

💡 Teacher's Secret Hint

Remember the relationship $pH + pOH = 14$ . If you find the concentration of $H^{+}$ , you can directly calculate pH. If you find ${OH}^{-}$ , calculate pOH first, then pH.

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Chemistry Question 131 – AP-EAMCET 2026

Clue 1 of 3: Acid-Base Neutralization

🧠 Full Solution Path