Question

Chromic acid, H2CrO4, is used in ceramic glazes and colored glass, and is particularly effective for cleaning insoluble organic residues off laboratory glassware. A 10.0 mL sample of chromic acid solution contains 42.0 mg chromium (which is found in the chromic acid. What volume (in mL) of 0.140 M KOH is needed to neutralize the 10.0 mL sample of chromic acid?

___?___mL KOH

Answer 1

Step-by-step explanation:

The balanced chemical equation for the neutralization reaction between chromic acid (H2CrO4) and potassium hydroxide (KOH) is:

H2CrO4 + 2KOH → K2CrO4 + 2H2O

From the equation, we see that 1 mole of H2CrO4 reacts with 2 moles of KOH.

First, we need to calculate the number of moles of chromium in the 10.0 mL sample of chromic acid:

mass of chromium = 42.0 mg

molar mass of chromium = 52.00 g/mol

moles of chromium = (42.0 mg / 1000 mg/g) / 52.00 g/mol = 0.000807 moles

Since the stoichiometry of the reaction is 1:2 for H2CrO4:KOH, we need twice as many moles of KOH as we have of H2CrO4 to fully neutralize the acid. Therefore, we need:

2 x 0.000807 = 0.001614 moles of KOH

Now we can use the molarity and the number of moles of KOH to calculate the volume of KOH solution needed:

moles of KOH = molarity x volume (in L)

volume (in L) = moles of KOH / molarity

volume (in L) = 0.001614 mol / 0.140 mol/L = 0.01153 L

Finally, we need to convert the volume from liters to milliliters:

volume (in mL) = 0.01153 L x 1000 mL/L = 11.53 mL

Therefore, the volume of 0.140 M KOH needed to neutralize the 10.0 mL sample of chromic acid is 11.53 mL (to three significant figures).