Equilibrium NEET 2006: Dilute HCl and Water Ionisation

What is the correct answer to the Equilibrium NEET 2006 dilute-HCl question?

B is the closest listed option for the Equilibrium NEET 2006 dilute-HCl question. The worked calculation gives a total hydrogen-ion concentration of approximately:

[H^{+}] \approx 1.05125 \times 10^{- 7} M

The task is to find the total equilibrium hydrogen-ion concentration in aqueous HCl at 298 K, given:

C = 10^{- 8} M, K_{w} = 10^{- 14}

The supplied choices are:

1.0 \times 10^{- 6} M

1.0525 \times 10^{- 7} M

9.525 \times 10^{- 8} M

1.0 \times 10^{- 8} M

The bank rates this question hard. Its 90-second expected solving time is a question-bank target, not an observed student average.

How do you count all the ions before calculating?

The official solution uses charge balance and water’s ionic product together. HCl dissociates completely, but that accounts only for the ions supplied by the acid. Water cannot be ignored: the added acid concentration is one-tenth of the hydrogen-ion concentration in pure water.

At 298 K, pure water has:

[H^{+}] = [O H^{-}] = \sqrt{K_{w}} = 10^{- 7} M

For the given HCl solution, complete dissociation establishes the chloride concentration:

H C l \to H^{+} + C l^{-}

[C l^{-}] = C = 10^{- 8} M

The solution must remain electrically neutral. Hydrogen ions supply the positive charge; chloride and hydroxide ions supply the negative charge:

[H^{+}] = [C l^{-}] + [O H^{-}] = C + [O H^{-}]

Water’s ionic product still applies after adding the acid:

[H^{+}] [O H^{-}] = K_{w}

[O H^{-}] = \frac{K_{w}}{[H^{+}]}

Define the total hydrogen-ion concentration and substitute: $h = [H^{+}]$

h = C + \frac{K_{w}}{h}

Do not add the pure-water concentration as a fixed contribution. Adding acid changes water’s equilibrium and suppresses its ionisation. The pure-water value is a reference, not an amount that remains unchanged in the acid solution.

How do you solve the quadratic and match option B?

Take the positive root of the charge-balance quadratic. It gives a concentration slightly different from the printed value in B, which is the closest option and the official answer. Do not alter the arithmetic to make them identical.

Multiply both sides by the total hydrogen-ion concentration:

h = C + \frac{K_{w}}{h}

$h^{2} = C h + K_{w}$ Rearrange and apply the quadratic formula: $h^{2} - C h - K_{w} = 0$

h = \frac{C \pm \sqrt{C^{2} + 4 K_{w}}}{2}

Because the ionic product is positive: $\sqrt{C^{2} + 4 K_{w}} > C$

The root with the minus sign gives a negative concentration. Reject it and retain:

h = \frac{C + \sqrt{C^{2} + 4 K_{w}}}{2}

Substitute the supplied values:

h = \frac{10^{- 8} + \sqrt{10^{- 16} + 4 \times 10^{- 14}}}{2}

h = \frac{10^{- 8} + \sqrt{4.01 \times 10^{- 14}}}{2}

Evaluate the square root:

\sqrt{4.01 \times 10^{- 14}} \approx 2.0025 \times 10^{- 7}

Hence:

h \approx \frac{10^{- 8} + 2.0025 \times 10^{- 7}}{2} = 1.05125 \times 10^{- 7} M

The printed option B is:

1.0525 \times 10^{- 7} M

These values are not identical. B is the closest listed choice, not the exact calculated result.

Check the direction of the change: $h > 10^{- 7} M$

The concentration is only slightly above that of pure water. This is consistent with adding a very small amount of strong acid.

Why does the wrong approximation give option D?

Option D comes from counting only acid-derived hydrogen ions. Complete dissociation tells you that HCl supplies its ions fully; it does not remove water from the equilibrium calculation.

The incorrect chain is:

HCl fully dissociates \Rightarrow [H^{+}] = C = 10^{- 8} M \Rightarrow option D

Test that proposed concentration using water’s ionic product:

[O H^{-}] = \frac{K_{w}}{h} = \frac{10^{- 14}}{10^{- 8}} = 10^{- 6} M

It cannot satisfy electroneutrality:

10^{- 8} \neq 10^{- 8} + 10^{- 6}

The same shortcut would give:

p H = - \log_{10} (10^{- 8}) = 8

That incorrectly describes this HCl solution as basic at 298 K. Use the acid-only approximation when:

C^{2} ≫ K_{w} or equivalently C ≫ \sqrt{K_{w}}

Here, that condition fails:

\frac{C^{2}}{K_{w}} = \frac{10^{- 16}}{10^{- 14}} = 10^{- 2}

Which three related questions check this method?

Calculate hydroxide concentration, find pH and test the concentrated-acid limit. These are original related practice questions, not additional verified NEET PYQs.

What is the hydroxide concentration in the same solution?

Subtract the acid concentration from the total hydrogen-ion concentration. Charge balance gives the answer without another quadratic: $[O H^{-}] = h - C$

[O H^{-}] \approx 1.05125 \times 10^{- 7} - 10^{- 8} = 9.5125 \times 10^{- 8} M

Cross-check with water’s ionic product:

[O H^{-}] = \frac{K_{w}}{h} = \frac{10^{- 14}}{1.05125 \times 10^{- 7}} \approx 9.5125 \times 10^{- 8} M

Both routes agree within rounding. Hydroxide concentration is slightly below its pure-water value, consistent with suppressed water ionisation after adding acid.

What is the pH of the original solution?

The pH is approximately 6.978, slightly below 7 rather than 8. Use the total equilibrium hydrogen-ion concentration, not the acid concentration.

With concentration expressed in molarity:

p H = - \log_{10} (1.05125 \times 10^{- 7})

p H = 7 - \log_{10} (1.05125) \approx 6.978

The solution is only slightly acidic, even though HCl is a strong acid. Acid strength describes dissociation, not how concentrated the solution is.

Can you ignore water for millimolar HCl at the same temperature?

Yes. At the same temperature and ionic product, the acid concentration is large enough that water’s contribution is negligible. Check the criterion before using the approximation: $C = 10^{- 3} M$

\frac{C^{2}}{K_{w}} = \frac{10^{- 6}}{10^{- 14}} = 10^{8} ≫ 1

Therefore:

[H^{+}] \approx C = 10^{- 3} M

[O H^{-}] \approx \frac{10^{- 14}}{10^{- 3}} = 10^{- 11} M

p H \approx - \log_{10} (10^{- 3}) = 3

The hydroxide term is tiny compared with the acid concentration. Dropping it from charge balance is therefore justified.

Same-chapter extension: Try Equilibrium NEET 2007: Combining Equilibrium Constants next. It practises equilibrium-constant manipulation rather than dilute-acid pH.

Frequently asked questions

What is the correct answer to the Equilibrium NEET 2006 HCl question?

B is the closest listed option and the official answer. For 10^-8 M HCl at 298 K, charge balance and Kw give [H+] approximately 1.05125 × 10^-7 M. The printed value in B, 1.0525 × 10^-7 M, is not identical to the calculated result.

Why is the pH of 10^-8 M HCl not 8?

Water contributes hydrogen ions and cannot be ignored at this low acid concentration. Using the total equilibrium [H+] of approximately 1.05125 × 10^-7 M gives pH approximately 6.978 at 298 K. Using only the HCl concentration incorrectly predicts a basic solution.

Can I add 10^-7 M from water to the HCl concentration?

No, because adding HCl suppresses water's ionisation, so its contribution does not remain fixed at the pure-water value. Instead, combine [H+] = C + [OH-] with [H+][OH-] = Kw. The positive root gives [H+] = (C + √(C² + 4Kw))/2.

When can I ignore water ionisation in HCl pH calculations?

Water's contribution is negligible when C² is much greater than Kw, or equivalently when C is much greater than √Kw. For 10^-3 M HCl at 298 K, C²/Kw = 10^8, so [H+] approximately equals 10^-3 M and pH approximately equals 3. This approximation fails for 10^-8 M HCl.