A 17.85 mL volume of nitric acid neutralizes 25.00 mL of 0.150 mol/L sodium hydroxide. What is the concentration of the nitric acid when the pH is exactly 7.0?

0.210 mol/L

Explanation:

NaOH (aq) + HNO3 (aq) = HOH + NaNO3 (aq)

25ml 17.85ml

6.150mol/L

NaOH = CxV = 0.150mol/L x 25 = 3.75mmol x 1/1 = 3.75mmol HNO3

C=n/V = 3.75mmol/17.85ml = 0.210 mol/L

0.005 M

Explanation:

Nitric acid is a strong acid, so it's concentration is equal to the amount of [H+] ions in solution. NaOH is also a strong base, meaning the same thing but with it's [OH-] ions. In order for a base to be neutralized, the number [H+] ions inserted must be equal to the number of [OH-] ions present.

Sodium Hydroxide Volume: 25 mL / 1000 = 0.025 L

Sodium Hydroxide Molarity: 0.15 mol/L * 0.025 L = 0.00375 M.

Now we can use M1V1=M2V2 to find the concentration of the nitric acid:

(0.025) (0.00375) = (17.85/1000) (X)

X = 0.005M