# What is the pH of #3.1 times 10^-3# #M# #HCl#?

##### 1 Answer

There are two ways you can do this. The easy way is to realize that

**EASY WAY**

Recall:

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

From the knowledge that

#color(blue)("pH") = -log(3.1 xx 10^(-3) "M") = 2.50_(8638306)#

#= color(blue)(2.51)#

**HARD WAY**

The hard way is to look up the

#"HCl"(aq) " "" " -> " "" ""H"^(+)(aq) + "Cl"^(-)(aq)#

#"I"" "3.1 xx 10^(-3)"M"" "" "" "" ""0 M"" "" ""0 M"#

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

#"E"" "(3.1 xx 10^(-3) -x)"M"" "" "x" "" "" "x#

#K_a = 1.3 xx 10^6 = x^2/(3.1 xx 10^(-3) - x)#

We could simply make the approximation that

#x^2 + K_ax - 3.1 xx 10^(-3)K_a = 0#

The quadratic formula gives, with

#x = 0.003099999993# ,

and we reject the root

Thus, we get that

#K_a = (3.099999993 xx 10^(-3))^2/(3.1 xx 10^(-3) - 3.099999993 xx 10^(-3))#

#= 1.37 xx 10^6 ~~ 1.3 xx 10^6# ,

as it was when we started, and

#["HCl"]_(eq) = 3.1 xx 10^(-3) - 3.099999993 xx 10^(-3) ~~ 0# ,

and

#["H"^(+)]_(eq) = 3.099999993 xx 10^(-3)# #"M"# #~~# #3.1 xx 10^(-3)# #"M"# ,

as expected. Finding ** exactly** zero... So, if we really wanted the "exact" answer:

#"pH" = -log["H"^(+)] = 2.50_(8638307) ~~ 2.51#

as before, only differing past the 8th decimal place (so who cares?).