30.03.2023 - 16:49

Preparation and standardization of a 1 M NaOH solution Calculate the following before beginning this task: a. The volume of a stock 6 M NaOH solution required to prepare 250 mL of a 1 M NaOH solution.

Preparation and standardization of a 1 M NaOH solution Calculate the following before beginning this task: a. The volume of a stock 6 M NaOH solution required to prepare 250 mL of a 1 M NaOH solution. b. The mass of potassium hydrogen phthalate (KHP) required to react with 25 mL of a 1 M NaOH solution. The molecular weight of KHP is 204.23 g/mol. c. The amount of water required to dissolve the quantity of KHP calculated in (b), given that the solubility of KHP is 10.2 grams per liter.

Answers (1)
  • djuliette
    April 8, 2023 в 02:26

    The problem asks for three calculations necessary to prepare and standardize a 1 M NaOH solution:

    a. To prepare 250 mL of a 1 M NaOH solution, we need to calculate the volume of a 6 M NaOH solution required. The equation to use is:

    M1V1 = M2V2

    Where M1 is the initial concentration of the solution, V1 is the initial volume of the solution, M2 is the final concentration of the solution, and V2 is the final volume of the solution.

    Plugging in the values, we get:

    (6 M) (V1) = (1 M) (250 mL)

    Solving for V1, we get:

    V1 = (1 M) (250 mL) / (6 M)

    V1 = 41.67 mL

    Therefore, we need to measure 41.67 mL of 6 M NaOH solution and add it to 208.33 mL of water to obtain 250 mL of a 1 M NaOH solution.

    b. The mass of potassium hydrogen phthalate (KHP) required to react with 25 mL of a 1 M NaOH solution can be calculated using the equation:

    Molarity = (moles of solute) / (liters of solution)

    Rearranging the equation to solve for moles of solute, we get:

    moles of solute = Molarity x liters of solution

    Plugging in the values, we get:

    moles of NaOH = (1 M) (0.025 L) = 0.025 moles of NaOH

    Since the balanced equation for the reaction between NaOH and KHP is:

    NaOH + KHP → NaKP + H2O

    We can see that the stoichiometry of the reaction is 1:1, which means that we need the same number of moles of KHP as NaOH. The molecular weight of KHP is 204.23 g/mol, so the mass of KHP required is:

    mass of KHP = 0.025 moles of KHP x 204.23 g/mol = 5.11 g

    Therefore, we need to measure 5.11 g of KHP.

    c. The amount of water required to dissolve the quantity of KHP calculated in (b), given that the solubility of KHP is 10.2 grams per liter.

    We have calculated that we need to dissolve 5.11 g of KHP. Since the solubility of KHP is given in grams per liter, we can calculate the volume of water required using the equation:

    Volume of water = Mass of solute / Solubility

    Plugging in the values, we get:

    Volume of water = 5.11 g / 10.2 g/L = 0.5 L

    Therefore, we need to dissolve 5.11 g of KHP in 500 mL of water.

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