Answer:
0.42 g
Explanation:
We have:
pH = 12.10 (25 °C)
V = 800.0 mL = 0.800 L
To find the mass of sodium hydroxide (NaOH) we can use the pH:
[tex] 14 = pH + pOH [/tex]
[tex]pOH = 14 - pH = 14 - 12.10 = 1.90[/tex]
[tex] pOH = -log ([OH^{-}]) [/tex]
[tex][OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M[/tex]
Now, we can find the number of moles (η) of OH:
[tex]\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles[/tex]
Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.
Finally, with the number of moles we can find the mass of NaOH:
[tex] m = \eta * M [/tex]
Where M is the molar mass of NaOH = 39.9 g/mol
[tex]m = 1.04 \cdot 10^{-2} moles * 39.9 g/mol = 0.42 g[/tex]
Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.
I hope it helps you!