# What is the pH of a solution of 750.0 mL of 0.500 M #HNO_2# to which 25.00 grams of sodium nitrite has been added (assume no volume change)?

##### 1 Answer

#### Answer:

#### Explanation:

The trick here is to recognize the fact that you're dealing with a **buffer solution** that contains nitrous acid, **weak acid**, and the nitrite anion, **conjugate base**, delivered to the solution by the soluble salt sodium nitrite.

The **Henderson - Hasselbalch equation**

#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#

Here

#"p"K_a = - log(K_a)#

and **acid dissociation constant** of the weak acid.

Nitrous acid has an acid dissociation constant equal to

#K_a = 4.0 * 10^(-4)#

which means that you can write the Henderson - Hasselbalch equation as

#"pH" = -log(4.0 * 10^(-4)) + log( (["NO"_2^(-)])/(["HNO"_2]))#

#"pH" = 3.40 + log( (["NO"_2^(-)])/(["HNO"_2]))#

Now, the problem wants you to assume that no volume change occurs when you add the salt, so you can say that

#["HNO"_2] = "0.500 M"#

In order to find the concentration of the nitrite anion, use the **molar mass** of *sodium nitrite* to calculate the number of moles of salt dissolved in the solution.

#25.00 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_2/(68.9953color(red)(cancel(color(black)("g")))) = "0.36234 moles NaNO"_2#

Sine sodium nitrite dissociates in a

#"NaNO"_ (2(aq)) -> "Na"_ ((aq))^(+) + "NO"_ (2(aq))^(-)#

you can say that the solution will contain **moles** of nitrite anions in

This implies that the **molarity** of the nitrite anions, which is calculated by dividing the number of moles of solute by the total volume of the solution **in liters**, will be

#["NO"_2^(-)] = "0.36234 moles"/(750.0 * 10^(-3)color(white)(.)"L") = "0.36234 moles"/0.750 * 1/"1 L"#

#= "0.48312 moles"/"1 L" = "0.48312 M"#

You can now plug this into the Henderson - Hasselbalch equation to find the

#"pH" = 3.40 + log( (0.48312 color(red)(cancel(color(black)("M"))))/(0.500color(red)(cancel(color(black)("M")))))#

#color(darkgreen)(ul(color(black)("pH" = 3.385)))#

The answer is rounded to three **decimal places**, the number of sig figs you have for the molarity of the nitrous acid.

Finally, does the result make sense?

Notice that you have

#["NO"_2^(-)] > ["NO"_2]#

This tells you that the solution contains more weak acid than conjugate base, which implies that the **lower** than the

In your case, you have

#"pH" = 3.385 < "p"K_a = 3.40" " -># makes sense.