!! LONG ANSWER !!
So, you're dealing with a buffer that contains monosodium phosphate, NaH_2PO_4NaH2PO4, and disodium phosphate, Na_2HPO_4Na2HPO4.
In aquous solution, the species that are of interest for your buffer are dihydrogen phosphate, H_2PO_4^(-)H2PO−4, which will act as a weak acid, and hydrogen phosphate, HPO_4^(2-)HPO2−4, which will act as the conjugate base.
Since nothing is actually added to the buffer, you have no net ionic equation to write. Now, because phosphoric acid is a triprotic acid, it will dissociate in three steps according to the following equilibrium reactions
H_3PO_(4(aq)) + H_2O_((l)) rightleftharpoons H_2PO_(4(aq))^(-) + H_3O_((aq))^(+)H3PO4(aq)+H2O(l)⇌H2PO−4(aq)+H3O+(aq), " "pK_(a1) pKa1
H_2PO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons HPO_(4(aq))^(2-) + H_3O_((aq))^(+)H2PO−4(aq)+H2O(l)⇌HPO2−4(aq)+H3O+(aq), " "pK_(a2) pKa2
HPO_(4(aq))^(2-) + H_2O_((l)) rightleftharpoons PO_(4(aq))^(3-) + H_3O_((aq))^(+)HPO2−4(aq)+H2O(l)⇌PO3−4(aq)+H3O+(aq), " "pK_(a3) pKa3
Notice that your buffer contains H_2PO_4^(-)H2PO−4 and HPO_4^(2-)HPO2−4, which means that the second equilibrium reaction will be established.
As a result, K_(a2)Ka2 will be the more important acid dissociation constant for your buffer.
Notice that the concentration of HPO_4^(2-)HPO2−4 is 2 times bigger than the concentration of H_2PO_4^(-)H2PO−4; this means that the dominant form of the acid will be HPO_4^(2-)HPO2−4 and that the solution's pH will be bigger than pK_(a2)pKa2.
Moreover, because the difference between K_(a2)Ka2 and K_(a3)Ka3 is so significant, the concentration of PO_4^(3-)PO3−4 can be neglected.
So, since this is a buffer, you can use the Henderson-Hasselbalch equation to solve for the pH. Since the second equilibrium is set, you'll use pK_(a2)pKa2, which is equal to
pK_(a2) = -log(K_(a2)) = -log(6.3 * 10^(-8)) = 7.20pKa2=−log(Ka2)=−log(6.3⋅10−8)=7.20
Therefore,
pH_"sol" = pK_(a2) + log(([HPO_4^(2-)])/([H_2PO_4^(-)]))pHsol=pKa2+log([HPO2−4][H2PO−4])
pH_"sol" = 7.20 + log((0.29cancel("M"))/(0.13cancel("M"))) = 7.20 + 0.35 = color(green)(7.55)
