Question

What is the maximum volume (in L) of 2.0 M HCl(aq) that the buffer prepared from 37.4 g of NH₄Cl and 1.25 L of 0.25 M NH₃ can tolerate without showing a pH change greater than 0.25 units? pK_a(NH₄⁺) = 9.25.

Answer 1

Initially without adding Hel) moles of Why al = 37:4 Molar mass NHuce = 53.5g/mol 53.5 = 0.699 - [Wh)= moly = 0.6099 volume - 1.25 l ume - 7.tal NH₂ - volume = 0.560 = 1. 25 l - [news] = 0.25M Initial pH i we know that Бам- ми aud = numele. pH = pka + dog (base bo = 9:25+ log 0.25 pH = 8.90 0.560 acid on adding Hill, pH will duercase as it is an for maximum 0.25 units change, New prt = 8.90 -0.25 = 8.65 Again. 8.65 = pka + log (base] 8.65 = 5.25 + log Cbase] I Cacid]

let's say Volume of He added = x litres : moles of Hill added = xx 2 moles B (rel=2 m) Total volume of solution = 195+& litres. moles of NH3 already = -28x0.25 = & moles جهان NH₂ t $ Hel 2x NHU I 0.699 Initial Change -2x 6+49 +24 -2x 5-24 ó 6.699+2x Final [ Nho e moles New volume 1225tax 5-2x TG 1:25+X [with ce} = 0.699 +2x 1.25ta 865 = 9.25 + log (-2) .699+2x) dog (ff-zx) (0.699+2x) = -0.60 7|X=0.0547 litres. Hence, we needo.o547 l of Hl.