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.
(a) 1.35
(b) 1.70
(c) 1.95
(d) 2.52
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) = ?
1.70
Ph determination
Please find ma an answer
1.7
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.
To calculate the pH of the resulting solution, we can follow these steps:
1. First, determine the moles of NaOH and HNO3 added to the solution.
- The moles of NaOH can be calculated using the formula: moles = concentration (molarity) × volume (in liters). In this case, the concentration is 0.100 M and the volume is 20.0 mL, which can be converted to liters by dividing by 1000: 20.0 mL ÷ 1000 = 0.020 L.
Moles of NaOH = 0.100 M × 0.020 L = 0.002 moles of NaOH.
- The moles of HNO3 can be calculated in a similar way. The concentration is 0.100 M and the volume is 30.0 mL, which is equivalent to 0.030 L.
Moles of HNO3 = 0.100 M × 0.030 L = 0.003 moles of HNO3.
2. Next, determine which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product formed. In this case, NaOH and HNO3 react in a 1:1 ratio, so the limiting reactant will be the one with fewer moles.
Since NaOH has 0.002 moles and HNO3 has 0.003 moles, NaOH is the limiting reactant.
3. The reaction between NaOH and HNO3 produces water and NaNO3. Since NaOH is the limiting reactant, all of it will react to form NaNO3.
4. Calculate the concentration of NaNO3 in the solution after the reaction. To do this, divide the moles of NaNO3 by the total volume of the resulting solution.
The total volume of the resulting solution can be obtained by adding the volumes of the initial solutions: 20.0 mL + 30.0 mL = 50.0 mL, which is equivalent to 0.050 L.
Concentration of NaNO3 = moles of NaNO3 / volume of resulting solution = 0.002 moles / 0.050 L = 0.040 M.
5. Finally, calculate the pH of the resulting solution using the concentration of NaNO3. Since NaNO3 is a salt of a strong acid (HNO3) and a strong base (NaOH), it dissociates completely in water, resulting in a neutral solution. Therefore, the pH of the resulting solution will be 7.
Therefore, none of the given options (a), (b), (c), or (d) are correct. The pH is 7.