The volume of potassium hydroxide solution required is 50 cm³
calculation
step 1
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
Answer:
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
