Ask Question
29 November, 01:27

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).

+1
Answers (2)
  1. 29 November, 02:18
    0
    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
  2. 29 November, 02:53
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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). ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers