# What is the #"pH"# for a solution containing #"0.500 M"# of a weak triprotic acid whose first #K_a# is #5.78 xx 10^(-10)#? Assume only the first proton is significant. Report your answer to two significant figures.

##### 1 Answer

The **ICE table** is the typical one for monoprotic weak acids, since we only look at the first proton here:

#"H"_3"A"(aq) + "H"_2"O"(l) rightleftharpoons "H"_2"A"^(-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" "0.500" "" "" "-" "" "" "0" "" "" "" "" "0#

#"C"" "-x" "" "" "-" "" "+x" "" "" "" "+x#

#"E"" "0.500-x" "-" "" "" "x" "" "" "" "" "x#

The **mass action expression** is:

#K_(a1) = (["H"_2"A"^(-)]["H"_3"O"^(+)])/(["H"_3"A"]) = 5.78 xx 10^(-10)#

#= x^2/(0.500 - x)#

This

#5.78 xx 10^(-10) ~~ x^2/0.500#

and so:

#["H"^(+)] = sqrt(5.78 xx 10^(-10) cdot 0.500) " M" = 1.7 xx 10^(-5)# #"M"#

and the

#"pH" = -log["H"^(+)] = 4.769#

To two sig figs it would then be:

#color(blue)("pH" = 4.8)#