Question

A chemist must prepare of sodium hydroxide solution with a pH of at . He will do this in three steps: Fill a volumetric flask about halfway with distilled water. Weigh out a small amount of solid sodium hydroxide and add it to the flask. Fill the flask to the mark with distilled water. Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step.

Answer 1

Answer:

0.0400 g for the example given below.

Explanation:

pH value is not provided, so we'll solve this problem in a general case and then we will use an example to justify it.

• By definition, $pH = -log[H_3O^+]$.
• NaOH is a strong base, as it's a hydroxide formed with a group 1A metal, so it dissociates fully in water by the equation: $NaOH (aq)\rightarrow Na^+ (aq) + OH^- (aq)$.
• From the equation above, using stoichiometry we can tell that the molarity of hydroxide is equal to the molarity of NaOH: $[NaOH] = [OH^-]$.
• Concentration of hydroxide is then equal to the ratio of moles of NaOH and the volume of the given solution. Moles themselves are equal to mass over molar mass, so we obtain: $[OH^-] = [NaOH] = \frac{n_{NaOH}}{V} = \frac{m_{NaOH}}{M_{NaOH}V}$.
• We also know that $pOH = 14.00 - pH = -log[NaOH]$. Take the antilog of both sides: $10^{-pOH} = 10^{pH - 14.00} = [NaOH] = \frac{m_{NaOH}}{M_{NaOH}V}$.
• Solve for the mass of NaOH: $m_{NaOH} = 10^{pH - 14.00}\cdot M_{NaOH}\cdot V$.

Now, let's say that pH is given as 12.00 and we use a 100-ml volumetric flask. Then we would obtain:

$m_{NaOH} = 10^{12.00 - 14.00}\cdot 39.997 g/mol\cdot 0.100 L = 0.0400 g$