Chemistry problem about pH.

Question

Can anyone help me with this problem? I can't get the answer, there must be something wrong in my reasoning, thank you so much!

 

"Calculate the pH of a solution obtained by adding 4 mL of nitric acid 85.7% (density=1.471 g/cm^3) to 2.50 L of pure water".

Answer 1

The steps involved are:

1.  Find out how much of the pure substance you have (HNO3) in 4 mL of HNO3 that is 85.7%.  This comes out to 3.428 mL of pure HNO3.

2.  Next, use the density of HNO3 to determine the amount of pure HNO3 you have in grams:  that comes out to be 5.0426 g. of HNO3.

3.  Next, find out how many moles of HNO3 you have by using the periodic table to calculate the gram-molecular weight and convert into moles of HNO3:  I get 0.080026 moles.

3.  Next, use the balanced reaction of HNO3 with water to produce the Hydronium ion:  I get the ratio of 1 mole of H3O+ is produced from one mole of HNO3, thus we have 0.080026 moles of H3O+ produced.

4.  Next, find the molar concentration of H3O+ in your solution (Molar concentration is the number of moles of solute you have divided by the total volume of the solution, in Liters--here the total volume of the solution is 2.50 liters of water plus the 4 mL of nitric acid.  We have then 2.504 liters of solution. We have then a 0.0319 M solution.

5.  Calculate the Ph using the formula:  Ph = -log[H3O+]    where the brackets indicate the Molar concentration of the H3O+.  I get:  1.496 for Ph.  Using significant digits, I round this off to:  1.5 for the Ph.

Answer 2

Calculate the pH of a solution obtained by adding 4 mL of nitric acid 85.7% (density=1.471 g/cm³) to 2.50 L of pure water.

 

Nitric acid is a strong acid, so if you can find the molarity of the HNO₃, then you've got the molarity of the H⁺ ion (also known as H₃O⁺ ion), because they will be one in the same.  HNO₃ doesn't exist at all in aqueous solution because the strong acid gives its H⁺ ion to water according to the following reaction:

 

HNO₃ + H₂O → H₃O⁺ + NO₃^–

 

For the problem, as stated, let's assume that the nitric acid concentration is "mass percent."  We can use the given data to calculate the molarity of the nitric acid (and of the H⁺ ion).

 

First, we'll need to calculate moles of HNO₃ from its mass in the solution, so let's calculate the molar mass of HNO₃.

 

molar mass = (1.008 x 1) + (14.01 x 1) + (16.00 x 3) = 63.018 g/mol.

 

Now, we need to distinguish between the 85.7% w/w HNO₃ solution and the compound HNO₃.  The solution is nitric acid, HNO₃ (aq), which is the molecule HNO₃ dissolved in water.  Let's just call the solution the "acid."

 

Let's calculate the number of moles of HNO₃ in our acid.  Pay careful attention to the units. We have the volume of acid, so we can start there and convert using the following plan.

 

mL acid → grams acid → grams HNO₃ → moles HNO₃

 

Note that the first conversion will involve the density of the acid.  The second conversion involves the mass percent of HNO₃ in the acid.  The third conversion involves the molar mass of HNO₃.  How do I know that?  I looked at the units and figured out how to get from one to the other.  The density, mass percent, and molar mass are all just conversion factors that get me from one unit to the other.

 

moles of HNO₃ = (4 mL acid) x (1.471 g acid / 1 mL acid) x (85.7 g HNO₃ / 100 g acid) x (1 mol HNO₃ / 63.018 g HNO₃) = 0.8 mol HNO₃

 

Now that we have the total moles of HNO₃ in the acid solution (which is the total moles of HNO₃ in both the concentrated solution (before adding water) and in the dilute solution (after adding water).  Why, because the water doesn't contain any HNO₃!

 

Total volume of water = [ (4 mL aqueous acid) x (1 L / 1000 mL) ] + 2.50 L = 2.50 L

 

Note: Thanks to sig figs, we can ignore the 4 mL of aqueous acid.  

(4 mL = 0.004 L; 2.50 L + 0.004 L = 2.50 L)

 

Now, the molarity is mol/L, so we can divide the total moles by the total liters.

 

molarity of HNO₃ in the final, diluted solution = (0.8 mol HNO₃) / (2.50 L) = 0.032 mol/L = 0.03 M

 

Therefore, the concentration of the hydrogen (or hydronium) ion is [H₃O⁺] = 0.03 M.

 

The pH is the negative of the base-10 logarithm of [H₃O⁺].

 

pH = – log [H₃O⁺] =  – log (0.03) = 1.5

 

Note: Because we have only one sig fig in the volume of concentrated acid, we can have only one sig fig in the pH.  A pH of 1.5 has one sig fig.  We don't count the number before the decimal point, because it has more to do with the power of 10 than the precision of the data.