show with calculations what would happen if 10mL of HCl and 10mL of NaOH were added to 40mL of H20 instead of?

Question

the buffers Na2HPO4 and Na3PO4

Answer 1

This question needs more information.

1) I'd need to know what the molarity of the HCl and NaOH solutions are. Knowing that would allow me to calculate the molarity of the HCl or NaOH solution after diluting each to 50 mL (that's the addition of the 40 mL of water).

2) I'd need more information on the buffer. I'd need enough to let me calculate how many moles of the two components there are in solution. I'd then calculate the moles of HCl (or NaOH) added and calculate the new moles of the two buffer components.

For example, suppose I add NaOH. That means that the HPO4^2- concentration would go down (it's the weak acid in the buffer) and the PO4^3- (it's the base) would go up. I'd calculate the amount down by subtraction of the moles of NaOH from the moles of HPO4^2-. That same amount of NaOH would be added to the PO4^3-

I'd recalculate the new molarities of the two buffer components (not really needed, but I usually teach it that way in class), then use the Henderson Hasselbalch equation to get the new pH of the buffer.

You will need the pKa for HPO4^2-

The reverse of the subtration and adding would be done with the HCl, then the new molarities and the H-H equation.

The behavior you are looking for is as follows:

a) adding the HCl to pure water will change the pH down by several pH units. Adding the NaOH to water would do the reverse, sending the pH up by several pH units.

b) adding HCl to the buffer will only change the pH by several tenths of a pH unit. This change will be in the more acid direction, that is, a lesser pH value. Adding the NaOH to the buffer will have the same magnitude effect (a change of only a few tenths of a pH unit) and the direction will be in the base direction (making a larger pH value).

HTH.

Answer 2

they would neutralise each other.