Determine the pH of a 7 × 10−5 M HNO3 solution. Your answer must be within ± 0.1%?

This concentration is large enough that

#"pH" = -log["H"_3"O"^(+)] = -log(7 xx 10^(-5)) = 4.15_(490196)#

This is clearly within #0.1%# of the exact solution:

#|(4.15490196 - 4.15490107)/(4.15490107)| xx 100% = 2.14 xx 10^(-5) %#

We can do it this way because in water, #["H"_3"O"^(+)] = 10^(-7) "M"# at #25^@ "C"#... and this is over #100# times bigger.

This could done using an ICE table for water... Why? Well, the concentration is very small.

#["HNO"_3] ~~ ["H"^(+)] = 7 xx 10^(-5) "M"#

This #"H"^(+)# interferes with the autoionization of water, which may or may not be significant.

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

#"I"" "-" "" "" "7 xx 10^(-5)" "" "" "0#
#"C"" "-" "" "+x" "" "" "" "" "+x#
#"E"" "-" "" "7 xx 10^(-5) + x" "" "x#

At #25^@ "C"#, #K_w = 10^(-14)#, so:

#K_w = 10^(-14) = ["H"_3"O"^(+)]["OH"^(-)]#

#= (7 xx 10^(-5)+x)x#

Compared to in water, #x# #"<<"# #7 xx 10^(-5)#, so we estimate that

#10^(-14) ~~ 7 xx 10^(-5)x#

#=> x = 1.43 xx 10^(-10) "M"#
(This is in fact, the exact solution to #x#.)

This is for #["OH"^(-)]#, but

#["H"^(+)] = 7 xx 10^(-5) + 1.43 xx 10^(-10) "M"#

Hence,

#color(blue)("pH" = 4.15)_(490107)#

where subscripts indicate digits past the reasonable digit of uncertainty.

This is as close as we could get to the exact solution.