Look at Kas and forget about NaCN i HCN! Such a buffer shows pH close to 10. You should neutralize NaCN completely and hydrogen phosphate partially.
OK. You meant that Using HNO3 0.0025 mole to neutralize CN- ?
because Kb CN- = 1.6x10^(-5) >> Ka HCN = 6.2x10^(-10) ?
So, we ignore hydrolysis effect of HCN?
for HPO4 2- ,let x =amount of HNO3
Since pH =7.21 --> [H+]= 6.16x10^-8 and Ka H2PO4- = 6.2x10^-8
Thus, 6.2x10^-8 =6.16x10^-8 [HPO42-]/[H2PO4-]
x = 2.492x10^-3 mole
all the needed amount of HNO3 = 4.992x10^-3 mole
So the final volume = 50 ml