50ml of 01m nh4oh 25ml nh4cl weere used to make a

Question

50ml of 0.1M NH₄OH ,25ml NH₄Cl weere used to make a buffer . what is the PH is if Pk_bis 4.8?

Solution) The given buffer is a base buffer

[NH4OH] =[(50*0.1)/75]

[NH4Cl] =[(25*2)/75]

From the principle V₁M₁=V₂M₂

For a base buffer p^OH = pK_b + log [Salt/Base]

thus p^OH 4.8+ log [{(25*2)/75}/{(50*0.1)/75}] = 4.8+ log (10) = 4.8+1 =5.8

Therefore p^H = (14-pOH) = (14-5.8) = 8.2