What is the mass of NaOH required to prepare 100.0mL of NaOH(aq) that
has a pH = 13.62 ?
A. 0.38g
B. 0.42g
C. 1.67g
D. 2.40x10^-14 g
OH- + H2O <-> H3O+ + O^-2
antilog-13.62 = 0.09976
#gNaOH=(0.1-0.09976)M x 40g/mol x 0.1L = 0.40g?
To calculate the mass of NaOH required to prepare 100.0 mL of NaOH(aq) with a pH of 13.62, we need to first calculate the concentration of hydroxide ions (OH-) in the solution.
The given pH value of 13.62 represents the concentration of hydronium ions (H3O+), which is equal to 10^-pH or 10^-13.62. Therefore, [H3O+] = 10^-13.62 = 0.09976 M.
Since NaOH dissociates completely in water, the concentration of hydroxide ions (OH-) is equal to the concentration of NaOH. Therefore, [OH-] = 0.09976 M.
To calculate the mass of NaOH needed, we can use the equation:
mass of NaOH = concentration of NaOH (M) x molar mass of NaOH (g/mol) x volume of solution (L)
Plugging in the values, we have:
mass of NaOH = 0.09976 M x 40 g/mol x 0.1 L = 0.39904 g
Rounding off to the appropriate number of significant figures, the mass of NaOH required is approximately 0.40 g.
Therefore, the correct option is A. 0.38 g.
To find the mass of NaOH required to prepare 100.0mL of NaOH(aq) with a pH of 13.62, you can use the equation:
OH- + H2O <-> H3O+ + O^-2
First, let's calculate the concentration of hydroxide ions (OH-) in the solution using the pH value. The pH is the negative logarithm (base 10) of the concentration of hydronium ions (H3O+).
To find the concentration of OH-, we can use the equation: [OH-] = 10^(-pOH)
Since pH = 13.62, the pOH would be 14 - 13.62 = 0.38 (pOH = 14 - pH).
Next, we can calculate the concentration of OH-. Taking the antilog of the pOH value, we get:
[OH-] = 10^(-0.38) = 0.09976 M
Now, let's calculate the mass of NaOH required. The formula weight of NaOH is 40 g/mol.
The equation to find the mass of NaOH is:
gNaOH = ([OH-] - [H3O+]) × molar mass of NaOH × volume of solution
Considering that the volume of the solution is 100.0mL, which is equivalent to 0.1L:
gNaOH = (0.09976 M) × 40 g/mol × 0.1 L = 0.39904 g
Rounding to the correct number of significant figures, the mass of NaOH required is approximately 0.40g.
So, the correct answer corresponding to the closest choice is A. 0.38g.
Genius up above me is wayyyy off answer is 1.67 g morons like him screw people over when they are they are trying to get help.
pH = 13.62
pOH = 0.38
(OH^-) = 0.417 M
M = moles/L; therefore,
moles = M x L = 0.417 x 0.1 L = 0.0417
grams = moles x molar mass = 0.0417 x 40 = ?? grams NaOH. I don't see 0.4.