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)?

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)?

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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

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.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years