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  1. Problems
  2. /What volume of 0.10 M sulphuric acid neutralises 36 mL of 0.15 M sodium hydroxide? …
MediumChemistryHigh School (9–10)GeneralAcids, Bases and Salts › pH and NeutralisationNumericalAPA-LevelIB

What volume of 0.10 M sulphuric acid neutralises 36 mL of 0.15 M sodium hydroxide? …

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

What volume of 0.10 M0.10\ \text{M}0.10 M sulphuric acid neutralises 36 mL36\ \text{mL}36 mL of 0.15 M0.15\ \text{M}0.15 M sodium hydroxide? H2SO4+2NaOH→Na2SO4+2H2O\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}H2​SO4​+2NaOH→Na2​SO4​+2H2​O.

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mL
Need a reminder? The key ideaShowHide

Moles = concentration (M) × volume (L); millimoles = M × mL. At neutralisation the moles of H+\text{H}^+H+ equal the moles of OH−\text{OH}^-OH−. For a one to one reaction, M1V1=M2V2M_1 V_1 = M_2 V_2M1​V1​=M2​V2​; otherwise use the ratio from the equation.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Base. 0.15 M×36 mL=5.4 mmol0.15\ \text{M} \times 36\ \text{mL} = 5.4\ \text{mmol}0.15 M×36 mL=5.4 mmol.

Step 2. Acid. 5.4÷2=2.7 mmol5.4 \div 2 = 2.7\ \text{mmol}5.4÷2=2.7 mmol.

Step 3. Volume. 2.7 mmol0.10 M=27 mL\dfrac{2.7\ \text{mmol}}{0.10\ \text{M}} = 27\ \text{mL}0.10 M2.7 mmol​=27 mL.

Answer: 27 mL27\ \text{mL}27 mL.

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