Question

Consider how best to prepare one liter of a buffer solution with
pH = **6.55** using one of the weak acid/conjugate
base systems shown here.

Weak Acid | Conjugate Base | K_{a} |
pK_{a} |
---|---|---|---|

HC |
C |
6.4 x 10 |
4.19 |

H |
HPO |
6.2 x 10 |
7.21 |

HCO |
CO |
4.8 x 10 |
10.32 |

How many grams of the **sodium** salt of the weak acid
must be combined with how many grams of the **sodium**
salt of its conjugate base, to produce **1.00** L of a
buffer that is **1.00** M in the weak base?

grams **sodium** salt of weak acid =

grams **sodium** salt of conjugate base =

Answer #1

This problem is similar to the one I answered earlier but with different data, so, use the same procedure but change the data. Here's the sample:

To determine the best buffer combination, calculate first the ratio (conjugate base/acid) which is calculated as:

ratio = 10^{pH-pKa}

The smallest number is the best buffer to use. The other way to
solve it, it's matching the pH value to the pKa closer to that pH.
In this case the
HPO_{4}^{2-}/H_{2}PO_{4}^{-}
with a pKa of 7.21 is the best for the buffer.

Now, as we know that the weak base has to be 1 M, then the acid
would have to be:

ratio =
[HPO_{4}^{2-}]/[H_{2}PO_{4}^{-}]

10^{6.55-7.21} =
[HPO_{4}^{2-}]/[H_{2}PO_{4}^{-}]

[HPO_{4}^{2-}]/[H_{2}PO_{4}^{-}]
= 0.2187 ------> [H_{2}PO_{4}] = 1 / 0.2188 =
4.57 M

Finally, the mass for these compounds would have to be:

mH_{2}PO_{4}^{-} = 4.57 * 1 * (23+2+31+64)
= 548.4 g

mHPO_{4}^{2-} = 1 * 1 * (2*23+1+31+64) = 142 g

Hope this helps

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Ka
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