How many grams of KHSO3 must be added to 250 mL of 0.1 M H2SO3 (Ka = 6 x 10^-8) to make a buffer with a pH of 7.00?

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The `K_(a1) ` of `H_2SO_3 ` is `1.54x10^-2` and is considerably stronger acid than the rest of the weak acids. However, the `K_(a2)` value of the second ionization is actually `6.73x10^-8` . I think there is something wrong with problem because we have to use the value of `K_(a1)` instead of `K_(a2)` .

However, if we use the `K_(a1)` , we can have the solution to be:

`pH = pK_a+ log ([sal t]/[acid])`

`7.00 = -log(1.54x10^-2)+ log ([sal t]/[0.1])`

`7.00 = 1.812 +log ([sal t]/[0.1])`

`7.00 - 1.812 =log ([sal t]/[0.1])`

`10^(5.1875) =[sal t]/[0.1]`

`153,992.65 = [sal t]/(0.1)` -> Somehow erroneous at this point because it is impossible to dissolve that amount in 250 mL of solution.

**Sources:**

**Molarity = moles/volume(liter)**

**Moles = molarity * volume**

Moles of H2SO3 = 0.1 M * (250/1000) L [ 1L = 1000 mL]

Moles of H2SO3 = 0.025 moles

And

**pKa = -log (ka)**

pKa of H2SO3 = -log (6*10^-8)=7.22

Now according to **Henderson Hasselbalch equation, **

**pH = pKa + log [salt]/[acid]**

pH = pka + log [KHSO3]/[H2SO3]

Here the pH is given as 7.0, plugging this value in the equation we get,

7.0 = 7.22 + log [KHSO3]/[H2SO3]

7.0-722 = log [KHSO3]-log [H2SO3] [log(a/b) = log a - log b]

-0.22 = log [KHSO3] - log [0.025]

log [KHSO3] = -0.22 + log [0.025] = -0.22 + (-1.60) = -1.82

[KHSO3] = 10^(-1.82) = 0.0151 mole

So the 0.0151 mol of KHSO3 is requried to make the pH of the buffer as 7.0

Now **moles = mass/molar mass**

**mass= moles * molar mass **

we have 0.0151 mol KHSO3 and its molar mass is 120.17 g/mol

Mass of KHSO3 = 0.0151 mol * (120.17 g/mol) = **1.18 g**

**Sources:**

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