Assuming complete reaction, what volume of 0.200 mol dm-3 potassium hydroxide solution (koh (aq)), is required to neutralize 25.0 cm3 of 0.200 mol dm-3 aqueous sulfuric acid, (h2so4 (aq))

The volume of potassium hydroxide solution required is 50 cm³

calculation

step 1

write the chemical equation

2KOH + H2SO4 → K2SO4 + 2H2O

step 2:find the number of moles of H2SO4

moles = vol in dm³x molarity (mol/dm³)

convert cm³ into dm³ = 25/1000=0.025 dm³

0.025 x 0.200 = 0.005 moles

step 3: use mole ratio to determine the moles of KOH

mole ratio of KOH : H2SO4 = 2:1 therefore the moles of KOH is 0.005 x2=0.010moles

step 4: find the volume of KOH

volume = moles/molarity

=0.010 / 0.2 = 0.050 dm³ = 0.050 x1000=50 cm³

50cm^3

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

2KOH + H2SO4 - > K2SO4 + 2H2O

From the balanced equation above, we obtained:

Mole of the acid (nA) = 1

Mole of the base (nB) = 2

Step 2:

Data obtained from the question include:

Concentration of the base, KOH (Cb) = 0.2moldm^-3

Volume of the base, KOH (Vb) = ?

Volume of the acid, H2SO4 (Va) = 25cm^3

Concentration of the acid, H2SO4 (Ca) = 0.2moldm^-3

Step 3:

Determination of the volume of potassium hydroxide (KOH) needed for the complete reaction. This is illustrated below:

Applying the formula CaVa/CbVb = nA/nB, the volume of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.2 x 25 / 0.2 x Vb = 1/2

Cross multiply to express in linear form as shown below:

0.2 x Vb = 0.2 x 25 x 2

Divide both side by 0.2

Vb = (0.2 x 25 x 2) / 0.2

Vb = 50cm^3

Therefore, the volume of the base, KOH required for the complete reaction is 50cm^3