Calculate the pH of the solution resulting from the addition of 20.0 mL of 0.100 M NaOH to 30.0 mL of 0.100 M HNO3.

Question

(a) 1.35
(b) 1.70
(c) 1.95
(d) 2.52

Answer 1

Ph determination

Answer 2

HNO3 + NaOH ==> NaNO + H2O

millimols HNO3 = 30 x 0.1 = 3.00
millimols NaOH = 20 x 0.1 = 2.00
millimols HNO3 in excess = 1.00

M HNO3 = mmols/mL = 1.00/50.0 = ?
Then pH = -log(HNO3) = ?

Answer 3

1.70

Answer 4

Please find ma an answer

Answer 5

1.7

Answer 6

My apologies for the confusion. Let's calculate the concentration of HNO3 first:

mol HNO3 = 3.00 x 0.001 = 0.003 mol
mol NaOH = 2.00 x 0.001 = 0.002 mol

Since NaOH is a strong base and HNO3 is a strong acid, they react in a 1:1 molar ratio. This means that all of the NaOH will react with an equal amount of HNO3, leaving 0.001 mol of HNO3 in excess.

To find the concentration of the excess HNO3, we need to subtract the amount reacted with NaOH from the initial amount of HNO3:

mol HNO3 in excess = 0.003 mol - 0.002 mol = 0.001 mol

Now we can calculate the concentration of the excess HNO3:

M HNO3 = mol HNO3 in excess / volume HNO3
= 0.001 mol / 0.030 L (30.0 ml)
= 0.0333 M

Finally, let's calculate the pH of the solution using the equation:

pH = -log[H+]

Since HNO3 is a strong acid, it fully dissociates in water. Therefore, the concentration of H+ ions is equal to the concentration of the excess HNO3. So,

pH = -log(0.0333)
= 1.48

The pH of the solution resulting from the addition of 20.0 mL of 0.100 M NaOH to 30.0 mL of 0.100 M HNO3 is approximately 1.48.

Therefore, the correct answer is not listed in the options provided.