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11 November, 05:01

A 0.24 m solution of a weak base a‑ is made. to a limited extent, the a‑ reacts with h2o to form some oh - and some of the corresponding weak acid ha. if the weak acid has a pka of 9.5, what will be the ph of the solution (to the nearest hundredths) ? for a hint see the acids and bases handout on canvas.

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  1. 11 November, 05:30
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    Solution: The Molarity of solution of weak base is 0.24 M and pK_{a} of weak acid formed by its reaction with water is 9.5.

    Let's initially all the weak base reacts with water to form hydroxide ion and weak acid as follows:

    A^{-}+H_{2}O/rightarrow OH^{-}+HA

    thus, concentration of A^{-}, OH^{-} and HA will become 0, 0.24 and 0.24 respectively.

    Reverse reaction will take place as follows:

    OH^{-}+HA/rightarrow A^{-}+H_{2}O

    The change in concentration takes place, let's the change be x thus, concentration of OH^{-}, HA and A^{-} will become 0.24-x, 0.24-x and x respectively.

    For the reaction, expression for acid dissociation constant will be:

    K_{a}=/frac{[A^{-}]}{[HA][OH^{-}]}

    Putting the values of concentration,

    K_{a}=/frac{[x]}{[0.24-x][0.24-x]} ... (1)

    From the pK_{a} of weak acid, acid dissociation constant can be calculated as follows:

    pK_{a}=-logK_{a}

    thus,

    K_{a}=10^{-pK_{a}}

    =10^{-9.5}

    =3.16/times 10^{-10}

    Putting the value in equation (1),

    3.16/times 10^{-10}=/frac{[x]}{[0.24-x][0.24-x]}

    Here, the value of K_{a} is very small so x can be neglected from denominator, thus,

    3.16/times 10^{-10}=/frac{[x]}{[0.24][0.24]}

    On solving,

    x=4.26/times 10^{-6}

    Concentration of hydroxide ion will be:

    [OH^{-}]=0.24-x=0.24-4.26/times 10^{-6}=0.2399

    now, pOH will be:

    pOH=-log[OH^{-}]

    =-log (0.2399)

    =0.62

    Use the following equation to calculate pH,

    pH+pOH=14

    thus,

    pH=14-pOH

    =14-0.62

    =13.38

    Thus, pH of solution is 13.38
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