How do you find the ph of .05M of H2C4H4O6 given Ka1=1.0 x 10 -3 and Ka2=4.6 x 10 -5 I am completely stuck.

Answer 1

You can see at http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base
How to derive the following equation
(where x=[H+])

x^2 -Kw -(Ca(Ka1x^2+2Ka12x) / (x^2+Ka1x+Ka12) =0

where Ca the concentration of the acid and Ka12 the equilibrium constant for the overall disociation reaction
Since for the overall reaction (2 H+ dissociating) Ka12=Ka1*Ka2,

x^2 -Kw -(Ca(Ka1x^2+2Ka1*Ka2x) / (x^2+Ka1x+Ka1Ka2) =0 =>
x^2-(10^-14) -( 0.05*(0.001x^2+ 2*(10^-3)*(4.6*10^-5)x) )/ (x^2+0.001x+ (10^-3)*(4.6*10^-5)) =0

Solve with the free on-line tool at
http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=advanced
x1= -0.0075<0 rejected
x2= -0.000092 <0 rejected
x3= -10^-13 <0 rejected
x4= 0.00663

pH =-log(0.00663) = 2.18