Question

23. A mixture contains 0.250 M benzoic acid, a monoprotic acid (Ka = 6.28 × 10‒5), and 0.400 M sodium benzoate. How many mL of a HCl solution whose pH is 0.523 should be added to 500 mL of this buffer to change its pH from what it is to pH = 4.20?

Answer 1

we have, 0.25N Aud (Bengoic auid) Salt(Sodiun Beng ata) 04M Ka = 6.28 No5 pka 4-202 pka= -1og Ka PH=Pka tosal TAcid O.25 M PH=4.4061 Nowj PH HCI = 0.523 ogLHt 0.523CH= 0.3M CHCIJ= CHt]0.3M 4.20 New pH 11 pH=pkat lag sab Cavid] 4.20-4.202 l Salt CAuid Baltl TAuid Con=moles Vo lume Asalt 0.9954 0.g954 naid

Reaction taking fare GHSCOOWA + HCI CHsCoo +Nacs trudually Aave we salt)-sau O.4 6 O.25 laid] naid Apter HCI additun we uant nsalt - 0.0.9954 naud we have Soo ml Quffer 00ML 025M auid 500 mlx O 25 mol 125mmol auit 500 ml e o.4 Salt 200 mmol Salt 5oomLK 0 4 mol

Say mmol o alt sredued and mmd dauid neased i.e from Asalt-n 0.9459 naud +n 0.9954 200-2 125 +2 200-2=1225 t o.9954x S elung or 31.815 mmol Thu, 37.815 mmd s Aegused uhih will eact

With 3to 875 mmol Salt to frdecs 37.875 mmol auid we tau CHerl0. 3M moles melarly Volune 31.315 mmd 0.3 Mmol L Volume VHCI=126.25 ML 26.25 ml e HC hequired to elstain equied pH.