# What is the pH of a solution in which #1\times10^(-7)# moles of the strong acid, HCl is added to one liter of water?

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*I am going to put what I think is the answer, please correct my solution if it's wrong*

finding molarity: #(1\cdot10^(-7)mol)/(1L)=1\cdot10^(-7)M#

R) #HCl\rightleftharpoons# ..... #H^++Cl-#

I) #1\cdot10^(-7)M# ...... 0 ....... 0

C) ...... #-x# ......... ... #x# ...... #x#

E) #1\cdot10^(-7)-x# .... #x# .... #x#

I don't know where to go from here as I don't know what #K_a# is.

I would apply the Henderson-Hasselbalch equation, but that's the next chapter down the line... and I don't have enough information for that anyway.

*I am going to put what I think is the answer, please correct my solution if it's wrong*

finding molarity:

R)

I)

C) ......

E)

I don't know where to go from here as I don't know what

I would apply the Henderson-Hasselbalch equation, but that's the next chapter down the line... and I don't have enough information for that anyway.

##### 1 Answer

#### Explanation:

Your starting point here will be the **auto-ionization of water**, which looks like this

#2"H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

At room temperature, water has an *ionization constant*,

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

Now, you know that hydrochloric acid is a **strong acid**, which implies that it ionizes completely in aqueous solution to produce hydronium cations

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

Notice that the concentration of hydronium cations is *equal* to the initial concentration of the acid, which in your case is

#(1 * 10^(-7)"moles HCl")/"1 L solution" = 1 * 10^(-7)color(white)(.)"M"#

The trick now is to realize that **after you add the moles of acid**, the auto-ionization equilibrium **still takes place!**

In other words, after you add

So you can say that after adding the acid, you have

#K_W = overbrace((x + 1 * 10^(-7)))^(color(blue)("the total concentration of H"_3"O"^(+))) * overbrace(" "x" ")^(color(purple)("the total concentration of OH"^(-))) = 10^(-14)#

This means that you have

#x^2 + 1 * 10^(-7)x - 10^(-14) = 0#

This quadratic equation will produce two values for *positive value* and a *negative value*. Since **concentration**, the negative value will hold no physical significance.

You will thus have

#x = 6.18 * 10^(-8)#

This means that the **total** concentration of hydronium cations in the final solution will be

#["H"_3"O"^(+)] = 6.18 * 10^(-8)"M" + 1 * 10^(-7)"M" = 1.618 * 10^(-7)"M"#

As you know, the pH of the solution is given by

#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#

In your case, you will have

#color(darkgreen)(ul(color(black)("pH" = - log(1.618 * 10^(-7)) = 6.8)))#

The value must be rounded to *one decimal place*, the number of **sig figs** you have for the number of moles of hydrogen chloride dissolved to make the solution of hydrochloric acid.

*Now, does the result make sense?*

Hydrochloric acid is a **strong acid**, so even in this very, very small concentration, the pH of the solution **must** come out to be