Question

If 0.106 g of chloroacetic acid (CH2ClCOOH) is diluted to 250.0 mL with distilled water, what volume of a 0.1200 M KOH standard must be added to reach the equivalence point?

Answer 1

To reach the equivalence point, 0.2805 mL of a 0.1200 M KOH solution must be added to 0.106 g of chloroacetic acid (CH2ClCOOH) diluted to 250.0 mL with distilled water.

Step-by-step explanation:

To determine the volume of a 0.1200 M KOH solution required to reach the equivalence point, we need to use stoichiometry. The balanced chemical equation for the reaction between chloroacetic acid (CH2ClCOOH) and KOH is:

CH2ClCOOH + KOH → CHClCOOK + H2O

From the equation, we can see that the ratio of CH2ClCOOH to KOH is 1:1. This means that for every mole of CH2ClCOOH, we need one mole of KOH. First, we need to calculate the number of moles of CH2ClCOOH in 0.106 g using the molar mass of CH2ClCOOH (94.51 g/mol). Then, we can use the stoichiometry to find the volume of 0.1200 M KOH solution needed.

Let's work through the calculations:

Number of moles of CH2ClCOOH = Mass / Molar mass = 0.106 g / 94.51 g/mol = 0.001122 mol

Volume of 0.1200 M KOH solution = Number of moles of CH2ClCOOH × Volume of CH2ClCOOH solution × (1 mol KOH / 1 mol CH2ClCOOH)

Now, substitute the values into the equation:

Volume of 0.1200 M KOH solution = 0.001122 mol × 0.2500 L × (1 mol KOH / 1 mol CH2ClCOOH) = 0.0002805 L = 0.2805 mL