# Question #e049b

##### 1 Answer

#### Answer:

#### Explanation:

The trick here is to realize that *hydrochloric acid*, **strong acid**, which implies that it ionizes **completely** in aqueous solution to form *hydronium cations*, *hydrogen ions*, *chloride anions*,

Since every molecule of hydrogen chloride contains **one atom** of hydrogen and **one atom** of chlorine, it follows that **one mole** of hydrochloric acid will ionize to form

,one moleof hydronium cations#1 xx "H"_3"O"^(+)# ,one moleof chloride anions#1 xx "Cl"^(-)#

You will thus have

#color(red)("H")"Cl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(color(red)(+)) + "Cl"_((aq))^(-)#

Use the *molarity* and *volume* of the initial solution to determine how many **moles** of hydrochloric acid it contains.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_(HCl) = "0.025 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#n_(HCl) = 3.75 * 10^(-4)"moles HCl"#

Since you know that hydrochloric acid ionizes in a

#n_(H_3O^(+)) = color(green)(|bar(ul(color(white)(a/a)3.8 * 10^(-4)"moles H"_3"O"^(+)color(white)(a/a)|)))#

The answer is rounded to two **sig figs**.

Now, you add **total volume** will now be

#V_"total" = "15 mL" + "25 mL" = "40. mL"#

Since this solution contains the **same number of moles** of hydronium cations as the initial solution, it follows that the concentration of the hydronium cations, which is **equivalent** to that of the hydrochloric acid, will be

#["H"_3"O"^(+)] = (3.75 * 10^(-4)"moles")/(40. * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.0094 M"color(white)(a/a)|)))#

Once again, the answer is rounded to two sig figs.

To find the pH of the solution, use the equation

#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#

Plug in the concentration of hydronium cations to get

#"pH" = - log(0.0094) = color(green)(|bar(ul(color(white)(a/a)2.03color(white)(a/a)|)))#