Calculate the volume of 0.200 M NH3 and 0.300 M NH4Cl to make a buffer solution having pH = 10.0?

Please help me to solve this problem. I don't know how to solve it

Calculate the volume of 0.200 M NH3 and 0.300 M NH4Cl to
make a buffer solution having pH = 10.0
Given: acid dissociation constant for NH4+ is 5.7x10-10.

Use the basic form of the Henderson-Hasselbalch equation:

pOH = pKb + log[BH+]/[B]

In this case, B = NH3, and BH+ = NH4+; NH4Cl produces NH4+ ions in solution.

Ka = 5.7*10^-10; pKa = 9.24 pKb = 14 - pKa = 4.76
To result in a pH of 10.0. that is a pOH of 14.0 - pH = 4.0

4.00 = 4.76 + log[[BH+]/[B]
[BH+]/[B] = 10^(4.00-4.7) = 0.175
If V1 is the volume of 0.200 M NH3 and V2 the volume of 0.300 M NH4Cl, then the total molarity of each is
[B] = 0.200*V1/(V1 + V2)
[BH+] = 0.300*V2/(V1 + V2)

[BH+/[B] = 0.300*V1/(0.200*V2)
0.175 = 1.5*(V1/V2)
V1/V2 = 0.117

If you need a total volume V of buffer, then V = V1 + V2 and you can solve for V1 and V2