[orthoformate]

A problem on the GRE chemistry practice test asks:

How many millimoles of K₂HP0₄ must be added to 100 mL of 0.100 M KH₂PO₄ solution to adjust the pH to 7.0

KH₂PO₄ K₂HP0₄  pKa= 5x10^-8

My first thought was to use the henderson hasselbalch eqn. pH = pKa + log (K₂HP0₄/KH₂PO₄)

7=6.3 + log (K₂HP0₄/KH₂PO₄)

0.7 = log (K₂HP0₄/KH₂PO₄)

10^0.7= (K₂HP0₄/KH₂PO₄)

(K₂HP0₄/KH₂PO₄)= 5.0

So in order for the pH to be 7, there must be 5 times as much (K₂HP0₄ with respect to KH₂PO₄) per liter

This is not in line with the answer they give, and I've been at it for a while. Does anyone have any advice?

Thanks!

[mjc123]

If Ka (not pKa) is 5 x 10^-8, pKa is not 6.3!

[orthoformate]

Thank you! I really appreciate your *delete me*

The new equation reads

7=7.3 + log (x/0.1 M)

-0.3 = log (x/0.1 M)

0.52 = (x/0.1 mols/L)

0.052 mols/L = x

so the answer is 5 mmol per 100 mL, which is in line with the correct answer on the sheet!