Question

A buffer solution is 0.409 M in H3PO4 and 0.258 M in KH2PO4. If Kal for H3PO4 is 7.5 x 10-3, what is the pH of this buffer solution? pH =

Answer 1

1) ka is not so small so we consider degree of dissociation x.

thus, to draw ICE table and find x using concentration and ka.

2) ka is very small, so not to consider degree of dissociation.

Date I_ l__ Sebo equilibeth to draw ICE table. 1₂PO4 + HO H₂PO4 + 0.409 M 0.258 0.409-X 0.258+4 worthing the equlibrium conotat equation, Ka - [h₂pow] [not] 0.258x + x² M₂ POy ? 0.409-1 x 25x10-3 x² +0.258x = -7.5*10-3x + 3.0675410 x 10.2 655 X 3.0675x103 x = -0.2655 € /(0.26551² + 4x3-0675x10 3 x=0.011 -(1,017 pha-loghhot) = -log(0.011) 11.967 Ch₂cool = 0.407 M. Cn. CooNa = 0.33/M. Ka = 1.87105 Cephes-log (1-9x1000)=4.9.444. Sincer tha io vely small so we can hegtect deepsee of dippseiction phapka + log [ch,000-] Tch, coon? 0-034 pH = 4.7 447+ log pH = 14.667 = 4.7447 Ooyot