What volume of a 0.12 m koh solution is needed to react with 23. ml of 0.22 m hcl? the reaction is: koh (aq) + hcl (aq) → kcl (aq) + h2o (l) ?

42.17 mL

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

KOH + HCl - > KCl + H2O

From the balanced equation above, the following were obtained:

Mole of the acid (nA) = 1

Mole of the base (nB) = 1

Step 2:

Data obtained from the question. This includes:

Molarity of the base (Mb) = 0.12 M

Volume of the base (Vb) = ?

Volume of the acid (Va) = 23 mL

Molarity of the acid (Ma) = 0.22 M

Step 3:

Determination of the volume of the base.

Applying the equation:

MaVa/MbVb = nA/nB

The volume of the base can be obtained as illustrated below:

MaVa/MbVb = nA/nB

0.22 x 23 / 0.12 x Vb = 1

Cross multiply to express in linear form.

0.12 x Vb = 0.22 x 23

Divide both side by 0.12

Vb = (0.22 x 23) / 0.12

Vb = 42.17 mL

Therefore, 42.17 mL of 0.12M KOH solution is needed to react with 23 ml of 0.22 M HCl.

The question is formatted wrong. The correct question should be

What volume of a 0.12 M KOH solution is needed to react with 23 ml of 0.22 M HCl? the reaction is: KOH (aq) + HCl (aq) → KCl (aq) + H2O (l) ?

Answer:-

42.2 mL

Explanation:-

Strength of HCl = 0.22 M

Volume of HCl = 23 mL = 23 / 1000 L = 0.023L

Number of moles of HCl = strength x volume

= 0.22 M x 0.023 L

= 0.00506 moles

According to the chemical equation

KOH (aq) + HCl (aq) → KCl (aq) + H2O (l)

1 mol of HCl reacts with 1 mol of KOH

0.00506 moles of HCl reacts with 0.00506 moles of KOH

Strength of KOH = 0.12 M

Volume of KOH = Number of moles / Strength

= 0.00506 moles / 0.12 M

= 0.0422 L

= 42.2 mL