In a titration, 100 mL of 0.026 M HCl (aq) is neutralized by 13 mL of KOH(aq). Calculate the molarity of KOH (aq).

Answers

Answer 1

Answer:

0.2M

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 100mL

Molarity of the acid (Ma) = 0.026 M

Volume of base (Vb) = 13mL

Molarity of the base (Mb) =..?

Step 2:

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

HCl + KOH —> KCl + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the molarity of the base, KOH. This can be obtained as follow:

MaVa/MbVb = nA/nB

0.026 x100 / Mb x 13 = 1

Cross multiply to express in linear form

Mb x 13 = 0.026 x 100

Divide both side by 13

Mb = 0.026 x 100 / 13

Mb = 0.2M

Therefore, the molarity of the base, KOH is 0.2M

Answer 2

Answer:

0.2M

Explanation:

KOH(aq) + HCl(aq) ⇒  KCl(aq) + H2O(l)

We express the moles of analyte (HCl) and titrant based (KOH) on their molar concentration:

M1 * V1 = M2 * V2

The molarity of the solution is calculated with the following equation:

M2 = V1 x M1 / ​​V2

Where:

V2 = valued sample volume

V1 = volume of titrant consumed (measured with the burette)

M1 = concentration of titrant solution

M2 = concentration of sample

M2 = 100mL * 0.026M / 13mL = 0.2M


Related Questions

18.35 mL of an HCN solution were titrated with 35.4mL of a 0.268M NaOH solution to reach the equivalence point. What is the molarity of the HCN solution

Answers

Answer:

0.517

Explanation:

HCN + NaOH → NaCN + H2O [balanced as written]

(35.4 mL) x (0.268 M NaOH) x (1 mol HCN / 1 mol NaOH) / (18.35 mL HCN) = 0.517 M HCN

Answer: 0.517

Explanation:

What are 3 stages of the water cycle are

Answers

Evaporation,condensation, and precipitation.

Two identical light bulbs are connected to a battery in a series circuit.
An ammeter is wired into the circuit at measures a current of the
battery to be 0.5 Amps. The two light bulbs are then wired in parallel.
The ammeter shows that the current:

Answers

Answer:

0.10 amps

Explanation:

HELP PLEASE ILL GIVE 25 pointsWhich of the following practices could help reduce erosion of water banks? a. buffer strips b. natural fertilizers and pesticides c. decrease in fossil fuel emissions d. all of the above Please select the best answer from the choices provided A B C D

Answers

Answer:

A. Buffer strips

Explanation:

The practice that could help reduce erosion of water banks is buffer strips.

What is erosion?

Erosion is the action of surface processes that removes soil, rock or dissolved material from one location on the Earth's crust, and then transports it to another location where it is deposited.

One of the practices that could be used to reduce the effect of erosion is buffer strips.

What buffer strips do is slow and filter storm runoff while helping to hold soil in place.

Learn more on buffer strips here; https://brainly.com/question/26872640

A sample of helium gas at room temperature is compressed from 100 cm3 to 20 cm3. Its new pressure is now 30 cm Hg. What was the original pressure of the gas?

Answers

Answer:

6 cm Hg

Explanation:

Boyles Law: P1V1=P2V2

(100 mL)(x)=(20 mL)(30 cm Hg)

x = 6 cm Hg

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