# 100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?

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100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with

pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?

100.0 mL of a pH=2.600 solution of hydrochloric acid is diluted with

pure water to a new volume of 200.0 mL. Calculate the pH of the new solution?

##### 2 Answers

#### Answer:

We use logarithms and powers of 10 to convert between pH numbers and concentrations of

The pH changes from 2.6 to 2.9. It moves closer to a neutral pH of 7.

#### Explanation:

pH is defined as 'minus the logarithm (in base 10) of the concentration of

That means the initial concentration of

If we double the amount of solvent (water) and leave the amount of solute (HCl) constant, the concentration of the solution will halve, because

The new concentration of

To find the pH of this final solution, we reverse the process. We take the logarithm (to base 10):

Then we change the sign back to positive, and the new pH is 2.90.

(Note: because HCl is a 'strong acid', I have assumed throughout that it is completely dissociated in solution)

#### Answer:

#### Explanation:

Here's a quick way of solving this problem. You know that the pH of the solution is given by the *concentration* of hydronium cations,

When you **double** the volume of the solution, you essentially **halve** the concentration of hydronium cations. This means that if you take

#["H"_ 3"O"^(+)]_ "dil" = (["H"_ 3"O"^(+)]_0)/2 -># the concentrationafter the dilution

You know that

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

You can thus say that **after** the solution is diluted, its pH will be

#"pH"_ "dil" = - log(["H"_ 3"O"^(+)]_"dil")#

This, in turn, is equivalent to

#"pH"_ "dil" = - log((["H"_ 3"O"^(+)]_0)/2)#

#"pH"_ "dil" = -[log(["H"_ 3"O"^(+)]_ 0) - log2]#

#"pH"_ "dil" = log2 - log(["H"_3"O"^(+)]_0)#

But since

#"pH"_ 0 = - log(["H"_ 3"O"^(+)]_0) = 2.600#

you will have

#"pH"_ "dil" = log2 + "pH"_ 0#

#color(green)(bar(ul(|color(white)(a/a)color(black)("pH" _ "dil" = 2.600 + log 2 = 2.901)color(white)(a/a)|)))#

The answer is rounded to three *decimal places*.

Finally, does this result make sense?

Diluting the solution means **decreasing** its concentration of hydronium cations, which in turn implies **increasing** the pH of the solution, i.e. making it *less acidic*.