- log(0.048 M NaOH)
= 1.3 pH
--------------need Molarity?
1/10 1.3
= 0.050 M H3O
---------------------
To prepare a 0.10 M NaOH solution, you need to dilute the 6.0 M solution. You can use the formula: M1V1 = M2V2 where M1 = 6.0 M, V1 is the volume of 6.0 M NaOH needed, M2 = 0.10 M, and V2 = 1000 mL (1.0 L). By substituting the values into the formula, you can calculate the volume of the 6.0 M NaOH needed to prepare the 0.10 M solution.
A 1.0 M hydrochloric acid (HCl) solution would have the largest concentration of hydronium ions (H3O+) since each mole of HCl dissociates completely to form one mole of H3O+. This means that the concentration of H3O+ ions in a 1.0 M HCl solution would also be 1.0 M.
- log(1 X 10 -12 M) = 12 pH -----------------------very little room for H3O
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
The pH of a solution with a H3O+ concentration of 7.9x10-11 M is approximately 10.1. This is because pH is calculated as -log[H3O+], so -log(7.9x10-11) ≈ 10.1.
This solution contain 26,3 g NaOH.
To prepare a 0.10 M NaOH solution, you need to dilute the 6.0 M solution. You can use the formula: M1V1 = M2V2 where M1 = 6.0 M, V1 is the volume of 6.0 M NaOH needed, M2 = 0.10 M, and V2 = 1000 mL (1.0 L). By substituting the values into the formula, you can calculate the volume of the 6.0 M NaOH needed to prepare the 0.10 M solution.
To calculate the pH after each volume of NaOH is added, you need to determine the moles of HBr initially, moles of NaOH added, and then use these values to calculate the final concentration of H3O+ ions in the solution. From there, you can calculate the pH using the equation pH = -log[H3O+]. Repeat these calculations for each volume of NaOH added to find the pH at 12.5 ml, 18.7 ml, 20.0 ml, 29.5 ml, and 34.2 ml.
A 1.0 M hydrochloric acid (HCl) solution would have the largest concentration of hydronium ions (H3O+) since each mole of HCl dissociates completely to form one mole of H3O+. This means that the concentration of H3O+ ions in a 1.0 M HCl solution would also be 1.0 M.
The reaction between HNO3 and NaOH is a 1:1 molar ratio. This means that the moles of HNO3 required to neutralize the NaOH is the same as the moles of NaOH. Given that 20.0 ml of HNO3 is needed to neutralize 10.0 ml of a 1.67 M NaOH solution, the molarity of the HNO3 solution is twice the molarity of the NaOH solution, which is 3.34 M.
- log(1 X 10 -12 M) = 12 pH -----------------------very little room for H3O
The pH of a solution with a H3O+ concentration of 7.9x10-11 M is approximately 10.1. This is because pH is calculated as -log[H3O+], so -log(7.9x10-11) ≈ 10.1.
The pH of a 0.33 M solution of NaOH is approximately 13.5. This is because NaOH is a strong base that dissociates completely in water to give hydroxide ions, which results in a highly alkaline solution.
By definition: pH = -log[H3O+]So pH = -log(7.4*10-9) = 8.13
3. since the [H+]=0.001 M then pH= -log[H+] -log(0.001)=3 pH=3.
Molarity = moles of solute/volume of solution 0.53 M NaOH = moles NaOH/3.8 Liters = 2.014 moles, or about 80 grams
-log(10^-4 M NaOH) = 4 14 - 4 = 10 pH NaOH -----------------------