How much NaOH (in grams) is needed to prepare 546 mL of solution with a pH of 10.00?
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Hmmm, well then
Considering that NaOH is a base, we must first find the pOH.
pH + pOH always = 14
In that case subtract the pH from 14
pOH = 14 - 10 = 4pOH
Now we must find the [OH-] concentration
Therefore we use the 10^-pOH formula
[OH-] = 10^-4 = 0.0001M
We now have the molarity of NaOH.
Now, multiply the molarity by the volume
Molarity is measured in Litres, therefore the volume has to be in litres too.
1L = 1000mL
0.0001 x 546 / 1000 = 5.46*10^-5mols of NaOH
Obtain the Mm of NaOH
Na = 22.989768
O = 15.9994
H = 1.00794
NaOH = 39.997108g/mol
Simply multiply the mols of NaOH by the Mm of NaOH
5.46*10^-5 x 39.997108 = 0.002g of NaOH
Considering that NaOH is a base, we must first find the pOH.
pH + pOH always = 14
In that case subtract the pH from 14
pOH = 14 - 10 = 4pOH
Now we must find the [OH-] concentration
Therefore we use the 10^-pOH formula
[OH-] = 10^-4 = 0.0001M
We now have the molarity of NaOH.
Now, multiply the molarity by the volume
Molarity is measured in Litres, therefore the volume has to be in litres too.
1L = 1000mL
0.0001 x 546 / 1000 = 5.46*10^-5mols of NaOH
Obtain the Mm of NaOH
Na = 22.989768
O = 15.9994
H = 1.00794
NaOH = 39.997108g/mol
Simply multiply the mols of NaOH by the Mm of NaOH
5.46*10^-5 x 39.997108 = 0.002g of NaOH
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Concentration is defined as n moles/L solution
pOH in particular is what we're interested in. Expanding the molarity equation helps make the objective of this problem more clear (finding the moles of NaOH, enabling us to find the grams of NaOH)
pOH = -log[OH-] where OH- is n moles/liters of solution
-log(x/.546L) = 4
Solve for x, which gives you approximately 5.46x10-4 moles of NaOH in this solution. Multiply this # by the molar mass of NaOH (40.0g) and you get .002184 grams (good luck weighing that out!)
pOH in particular is what we're interested in. Expanding the molarity equation helps make the objective of this problem more clear (finding the moles of NaOH, enabling us to find the grams of NaOH)
pOH = -log[OH-] where OH- is n moles/liters of solution
-log(x/.546L) = 4
Solve for x, which gives you approximately 5.46x10-4 moles of NaOH in this solution. Multiply this # by the molar mass of NaOH (40.0g) and you get .002184 grams (good luck weighing that out!)