Question

4 points 3. What is the pH of the buffer resulting of the mixing of 34,4 g of sodium dihydrogen phosphate (NaH2PO4) and 48.5 g of sodium hydrogen phosphate (Na2HPO4) in 2.5 L of distilled water? Your answer

Answer 1

To calculate pH of a buffer you need to use the Henderson Hasselbalch equation:

(Salt] pH = pka + log [Acid]

K_a is acid dissociation constant which corresponds to the acid dissociation of : H₂PO₄^- → HPO₄^2- + H⁺

Ka for that = 6.3 x 10^-8 . pKa = -log( 6.3 x 10^-8) = 7.2

Here,salt = HPO₄^2- = Na₂HPO₄ and acid = H₂PO₄^- = NaH₂PO₄

[…] denotes concentration in molarity.

Now, Molarity = Number of moles/Volume in L.

Both the salt and acid are mixed in the same container, so they will have same volume ( 2.5 L).

Thus, when taking the ratio of their molarities, the volume will cancel out.

Thus, effectively, [Salt]/[Acid] = ( number of moles of Na₂HPO₄  ) /( number of moles of NaH₂PO₄)

The mass in grams of the compounds is given. To convert grams to moles, use :

Number of moles = Mass ( in g) / Molar mass

The molar masses are - Na₂HPO₄ = 142 g/mol and NaH₂PO₄ = 120 g/mol

So, number of moles of Na₂HPO₄ = (48.5 g)/(142 g/mol) = 0.342 moles

And number of moles of NaH₂PO₄= ( 34.4 g)/(120 g/mol) = 0.287 moles

So, putting the values :

[0.342] pH = 7.2 + log [0.287]

Solving, pH = 7.28 ( answer)