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100mL of a buffer solution contains 0.1M...

`100mL` of a buffer solution contains `0.1M` each of weak acid `HA` and salt `NaA`. How many gram of `NaOH` should be added to the buffer so that it `pH` will be `6`? `(K_(a)` of `HA = 10^(-5))`.

A

`0.328`

B

`0458`

C

`4.19`

D

None

Text Solution

Verified by Experts

The correct Answer is:
A
For acidic buffer, `pH = pK_(a) + "log" (0.1)/(0.1)`
`pH = pK_(a) = - log (10^(-5)) = 5`.
Rule: `ABC` (In acidic buffer `(A)`, on addition of `S_(B)(B)`, the concentration of `W_(A)(A)` decreases and that of salt increases.
Let `x M` of `NaOH` is added.
`pH_(new) = 5 + log((0.1 +x)/(0.1-x))`
`6-5 = log ((0.1 +x)/(0.1 - x))`
`((0.1 +x)/(0.1 - x)) = Antilog (1) = 10`
Solve for `x:`
`x = 0.082 M = (0.082)/(1000) xx 100`
`= 0.0082 mol (100 mL)^(-1)`
`= 0.0082 xx 40 g(100 mL)^(-1)`
`= 0.328 g`
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Knowledge Check

  • 100 mL of a buffer solution contains 0.1 M each of weak acid HA and salt NaA . How many gram of NaOH should be added to the buffer so that it pH will be 6 ? ( K_(a) of HA=10^(-5) ).

    A
    `4.19`
    B
    `0458`
    C
    `0.328`
    D
    None

  • How many grams of NaOH should be added to water to prepare 250 ml solution of 2 M NaOH?

    A
    `9.6 xx 10^(3)`
    B
    `2.4 xx 10^(3)`
    C
    20
    D
    24

  • What is the pH of 1.00 L sample of a buffer solution containing 0.10 mol of benzoic acid and 0.10 mol of sodium benzoate of which 0.010 mol of NaOH has been added ? [ K_a benzoic acid= 6.5xx10^(-5) ]

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