Question

A solution of potassium hydroxide (KOH) was titrated against a solution of hydrochloric acid. It took 35cm3 of the hydrochloric acid to completely neutralise the potassium hydroxide. Work out how many moles of acid there were in this volume if the concentration of the acid was 2mol/dm³

Answer 1

The number of moles of HCl used to neutralize the KOH solution is calculated by multiplying the concentration of HCl (2 mol/dm³) by the volume in dm³ (0.035 dm³), resulting in 0.07 moles of HCl.

Step-by-step explanation:

To determine the number of moles of hydrochloric acid used to neutralize the potassium hydroxide solution, we use the equation:

Moles = Concentration × Volume

Where concentration is given in moles per cubic decimeter (mol/dm³), and volume is given in cubic decimeters (dm³). First, we need to convert the volume from cm³ to dm³:

35 cm³ = 35 / 1000 dm³ = 0.035 dm³

Now, applying the formula:

Moles = 2 mol/dm³ × 0.035 dm³

Moles = 0.07 mol

So there were 0.07 moles of HCl in the 35 cm³ volume used to titrate the KOH solution.

Answer 2

The number of moles of potassium hydroxide that would react with 1 mole of hydrochloric acid will be 1.

Stoichiometric Ratio

From the balanced equation of the reaction as shown below:

One would see that the stoichiometric mole ratio of to that of is 1:1.

This means that 1 mole of KOH will require 1 mole of HCl for a complete reaction that obeys the law of conservation.

Step-by-step explanation: