You have 200 mL of 1 M ammonia solution (pKa=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.25? Answer: 100ml

Guys please help me.

I know how to do the question when there is 100 ml of 1m ammonia solution. But now the question is 200 mL and I am getting confused. Can someone please show me the steps? Thank you

This is the solution for when it's 100 ml of 1m ammonia. I looked at it and I did myself and got it right.

"So we know that from the HH equation, 9.5 = 9.25 + log (NH3)/(NH4+)

Therefore log (NH3/NH4) = 0.25, use a calculator to find that log x =0.25, where x = 1.78, the ratio of NH3/NH4+.

NH3 ratio to start is 0.1, so basically since HCl completely disassociates (and in a 1 H+ to HCl manner) we know that any HCl added will form the conjugate acid NH4+. If we represent the amount of HCl added as x, we know that NH3 will be 0.1 - x at the buffer solution (starting ammonia will react with the HCl added). We know that x will also be the amount of NH4+ formed.

Therefore we can represent NH3/NH4+ as (0.1-x)/x, we calculated this ratio as 1.78 earlier. Simply solve for x to get the amount of HCl added (also the amount of NH4+ formed)."