Exams › NEET › Chemistry

What is the [OH⁻] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂?

1. 0.40 M
2. 0.005 M
3. 0.12 M
4. 0.10 M

Correct answer: 0.12 M

Solution

Ba(OH)₂ provides 2 OH⁻ ions per molecule. Moles of OH⁻ = 30.0 mL × 0.10 M × 2 = 0.006 mol. Moles of H⁺ = 20.0 mL × 0.050 M = 0.001 mol. Excess OH⁻ = 0.006 - 0.001 = 0.005 mol. Total volume = 50.0 mL = 0.050 L. [OH⁻] = 0.005 mol / 0.050 L = 0.10 M.

Related NEET Chemistry questions

• A beaker contains a saturated solution of copper(l) chloride. Copper(l) chloride is slightly soluble salt with a solubility product of \( 1.2 \times 10^{-6} \) Which of the following salts when added to the solution would precipitate copper(l) chloride?
• The function of 'buffer solution' is:
• When \( C a(O H)_{2} \) attains solubility equilibrium, the:
• Predict if there will be any precipitate by mixing \( 50 \mathrm{mL} \) of \( 0.01 \mathrm{M} \) NaCl and 50 mL of \( 0.01 \mathrm{M} A g N O_{3} \) solution. The solubility product of \( A g C l \) is \( 1.5 \times \) \( 10^{-10} \)
• Calculate the ratio of \( H C O O^{-} \) and \( F^{-} \) in a mixture of \( 0.2 M \) HCOOH \( \left(K_{a}=\right. \) \( \left.2 \times 10^{-4}\right) \) and \( 0.1 M H F\left(K_{a}=6.6 \times\right. \) \( \left.10^{-4}\right) \)
• In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?

⚔️ Practice NEET Chemistry free + battle 1v1 →