Question #0ce76

You know that hydrochloric acid and sodium hydroxide neutralize each other in a #1:1# mole ratio

#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))#

so start by calculating the number of moles of each reactant.

#70.0 color(red)(cancel(color(black)("mL"))) * "4.00 moles HCl"/(10^3color(red)(cancel(color(black)("mL")))) = "0.280 moles HCl"#

#30.0 color(red)(cancel(color(black)("mL"))) * "8 moles NaOH"/(10^3color(red)(cancel(color(black)("mL")))) = "0.240 moles NaOH"#

Since you have fewer moles of sodium hydroxide than the number of moles needed to ensure that all the moles of hydrochloric acid take part in the reaction, you can say that the sodium hydroxide will act as a limiting reagent.

This implies that the base will be completely consumed before all the moles of hydrochloric acid will get the chance to react #-># you don't have a complete neutralization.

The reaction will consume #0.240# moles of sodium hydroxide and #0.240# moles of hydrochloric acid and leave you with

#"0.280 moles " - " 0.240 moles = 0.0400 moles HCl"#

The total volume of the resulting solution will be

#"70.0 mL + 30.0 mL = 100.0 mL"#

Now, hydrochloric acid ionizes in a #1:1# mole ratio to produce hydronium cations, so you can say that you have

#["H"_3"O"^(+)] = "0.0400 moles"/(100.0 * 10^(-3)color(white)(.)"L") = "0.400 M"#

As you know, the #"pH"# of a solution is defined as

#"pH" = -log(["H"_3"O"^(+)])#

In your case, this will be equal to

#color(darkgreen)(ul(color(black)("pH" = -log(0.400) = 0.40)))#

I'll leave the answer rounded to two decimal places, but you could have it rounded to three decimal places

#"pH" = 0.398#

because you have three sig figs for your values.

Given:
#70.0ml(4.00M HCl) + 30.0ml(8.00M NaOH)# =>

... because moles = Molarity x Volume(L) ...

=> #[(0.070)(4.00)]"mole" HCl + [(0.30)(8.00)"mole" NaOH]#
=> #0.28 "mole" HCl + 0.24"mole" NaOH#

... because reaction ratio for HCl + NaOH => NaCl + HOH is a 1:1 rxn ratio, then NaOH is the limiting reagent and only 0.24 mole of the available 0.28 mole of HCl will be consumed and converted into NaCl ... The excess HCl will dictate the pH of the final mix.

=> #(0.28 - 0.24)"mole" HCl# remains in excess at equivalence point.
=> #0.04"mole"HCl# in excess at equivalence point.

... moles HCl (excess) = moles #H^+# because HCl is strong acid and ionizes 100%.

... to calculate pH = log [#H^+#], the concentration of #H^+# in #(("moles"H^+)/("Liter""Solution"))# needs to be determined.

=> #[H^+] = (0.04"mole" H^+)/((0.070 + 0.030)Liter Solution) = 0.40M HCl#

=> #pH = -log[H^+] = -log(0.40) = -(-0.40) = 0.40#