Respuesta :
Answer:
b. 10 mL
Explanation:
First we calculate the amount of H⁺ moles in the acid:
- pH = -log [H⁺]
- [H⁺] = [tex]10^{-pH}[/tex]
- [H⁺] = 10⁻⁵ = 1x10⁻⁵M
100 mL ⇒ 100 / 1000 = 0.100 L
- 1x10⁻⁵M * 0.100 L = 1x10⁻⁶ mol H⁺
In order to have a neutral solution we would need the same amount of OH⁻ moles.
We can use the pOH value of the strong base:
- pOH = 14 - pH
- pOH = 14 - 10 = 4
Then we calculate the molar concentration of the OH⁻ species in the basic solution:
- pOH = -log [OH⁻]
- [OH⁻] = [tex]10^{-pOH}[/tex] = 1x10⁻⁴ M
If we use 10 mL of the basic solution the number of OH⁻ would be:
10 mL ⇒ 10 / 1000 = 0.010 L
- 1x10⁻⁴ M * 0.010 L = 1x10⁻⁶ mol OH⁻
It would be equal to the moles of H⁺ so the answer is b.
Answer:
We have to add 10 mL of base ( option B is correct)
Explanation:
Step 1: Data given
Volume of the strong acid = 100 mL = 0.100 L
pH = 5
pH of the strong base = 10
Step 2: Calculate molarity of the strong acid
pH =[H+] = 5
[H+]= 10^-5 M
Step 3: Calculate moles of the strong acid
Moles = molarity * volume
Moles = 10^-5 M * 0.100 L
Moles = 10^-6 moles
Step 4: Calculate pOH
pOH = 14 - 10 = 4
Step 5: Calculate [OH-]
[OH-] = 10^-4 M
Step 6: Calculate volume need
We need 10^-6 moles of base
Volume = moles / molarity
Volume = 10^-6 moles / 10^-4 M
Volume = 0.01 L
Volume = 10 mL
We have to add 10 mL of base ( option B is correct)