Recent content by emilj

oh okay thank you very much for your help :) i appreciate it!

so then if i sub that in, so.. 0.00564 M/0.2M =0.0282L, is that then correct?

oh that should be 0.2M not 0.02. my mistake

i've just had a look at the question again, i think I've calcualted the number of moles incorrectly, moles of ammonia should actually be, n= 0.1M/1.0L n=0.1 is that correct?

okay, so i began by finding the moles of ammonia in 1L, first found moles in 100mL c=n/v so n=cV n=0.100M/0.1 = 0.01 then using henderson hasselbach, pH=pKa+ log (NH4+/NH3) 9.5=9.25+log(NH4+/0.01) 0.25=log(NH4/0.01) 10^-0.25= (NH4/0.01) NH4= 0.564 x 0.01 n of NH4 = 0.00565...

oh ok, thank you for your feedback. is there any chance you would mind showing me the working for the question? im just a bit confused about which ratio you're talking about, I've completed the question and gotten an answer of 0.2823L which i know isn't right.

1. A buffer is prepared by mixing a 100.0ml of a 0.100M NH3 solution with a 0.200M solution of NH4Cl and making the total volume up to 1.000L of water. What is the volume of the NH4Cl solution required to achieve a buffer at pH=9.5 Ka of NH4+ = 5.6x10^-10 The textbook tell me that the...