If 1 volume of 0.1 KH2PO4 is mixed with 2 volumes of 0.1 OHEtNH2 (pka=9.44), what will be the pH of the mixture?
I know I know:
HH equation
I would do it this way.
.tNH2+H2PO4^-==>tNH3^+ + HPO4^-2
I assume the units you omitted on the concns are M. Make up a volume, say 100 mL for the phosphate and 200 mL for the amine. That gives, in millimoles.
begin 20.....10......0......0
change -10...-10.....+10.....+10
final..10.....0.....+10......+10
Now substitute into the HH equation. Won't that be a pH of 9.44?
pH = 9.44 + log(10/10)
Ur the BEST DR.BOB!
To calculate the pH of the mixture, you can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log [A-] / [HA]
where pH is the desired pH value, pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid is KH2PO4 and the conjugate base is H2PO4-. The pKa of H2PO4- is not provided, but you can find it in chemical literature or online databases. Once you have the pKa, you can plug in the values to calculate the pH.
To find the concentrations of the acid and the conjugate base, you need to use the information given in the question. It states that 1 volume of 0.1 M KH2PO4 is mixed with 2 volumes of 0.1 M OHEtNH2. Since the volumes are given, you can assume a total volume of 3 mL (1 mL from KH2PO4 and 2 mL from OHEtNH2).
First, let's calculate the moles of KH2PO4 and OHEtNH2:
Moles of KH2PO4 = Volume of KH2PO4 (in L) x Concentration of KH2PO4 (in M)
= 0.001 L x 0.1 M
= 0.0001 mol
Moles of OHEtNH2 = Volume of OHEtNH2 (in L) x Concentration of OHEtNH2 (in M)
= 0.002 L x 0.1 M
= 0.0002 mol
The KH2PO4 reacts with OHEtNH2 in a 1:1 ratio, so the moles of H2PO4- formed (the conjugate base) will be equal to the moles of KH2PO4 used. Therefore, the concentration of H2PO4- will be the same as the concentration of KH2PO4.
Now, let's calculate the concentration of H2PO4-:
Concentration of H2PO4- = Moles of H2PO4- / Total volume (in L)
= 0.0001 mol / 0.003 L
= 0.0333 M
With the concentration of H2PO4- known, you can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log [A-] / [HA]
= pKa + log (0.0333 M) / (0.0333 M)
Simplifying:
pH = pKa + log 1
Since the logarithm of 1 is 0, you have:
pH = pKa
So, the pH of the mixture will be equal to the pKa of H2PO4-.