Question

What is the maximum volume of 1.6 M NaOH(aq) that the buffer (pH=5.75) prepared from sodium acetate and 2.5 L of 0.55 M acetic acid can tolerate without showing a pH change greater than 0.25 units? pK_a of acetic acid is 4.75.

L

The number of significant digits is set to 2; the tolerance is +/-4%

Answer 1

PH₂ ( after addug NaOH ) = pkatlog 18 ( 1.375-1-62 ) = 13.75 + 1.6n . CH3COOH Butter solution of CH₃COON Oona and 2.5 l 0.55M PH=5.75 ; pka (CH₂COOH) = 4.75 Henderen equation pH = pka, kg 5.75=4.75 +669 (szet New De [MS Vs] {Nava] 1 = los NAVA (Agil NSVS Nava. los 10 = ng NSUS 10 Added Ла он Nava => NSVS= 10 x Nava =10 y 0-55X2155 = 13.75 meles salt reacts with CH₂ CoolH to form salt CH₂C00 na a) As a result concentrat con salt inc geages and aud decreases, 2-51X0.55=1-375 CH3 Coot +NaOH - CH₂COo NatH2O 1-375 1 6 X X (1) 13.75 -1.60 -1.60 Final 1-375-761 13-75716n. PHI (befor adding Nool) = pka flag 1.375 13-75 1375-1-64 =) PH₂-PH, = -log # log!3754167 1-375-1.60 - log10 + log 13-757161 +1-60 Initial Change 13-75 1375 0-25 7375-1.60 1-3757.60 0.25=-14 log (131 log 13.7577.69 1-25= 1375+16n log 18 18= log (1.375 16x! 13.75+1.61 1375-1.60 ره 28.89 24.75-tason 24.75~13-75 Il 13.7541.60 1.60+ 28.8n. -- 30.4n. IL -0.3618 30-4 = 0.362 l = 362 ml (NGOH)