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9 February, 00:05

Assuming complete reaction, what volume of 0.200 mol dm-3 potassium hydroxide solution (koh (aq)), is required to neutralize 25.0 cm3 of 0.200 mol dm-3 aqueous sulfuric acid, (h2so4 (aq))

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  1. 9 February, 00:21
    0
    The volume of potassium hydroxide solution required is 50 cm³

    calculation

    step 1

    write the chemical equation

    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
  2. 9 February, 00:31
    0
    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
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