. 12.75 grams of KOH are diluted to 125.0 ml with water. 10.0 ml of the resultin

Question

. 12.75 grams of KOH are diluted to 125.0 ml with water. 10.0 ml of the resulting solution are removed and diluted to a final volume of 250.0 ml. What is the concentration of KOH in the final solution?

2. A 25.0 ml sample of 0.0750 M solution of Ba(OH)2 is titrated with 0.200 M HCl. How many milliliters of the HCl solution are required to reach the equivalence point of the titration?

3. For each of the weak acids listed below, a) write a balanced equilibrium equation showing the dissociation and b) write the equilibrium constant expression and c) give the Ka value (from your lecture note manual). HClO2 (aq) HC7H5O2 (aq) (also written as, C6H5COOH) HIO4 (aq)

4. A 150.0 ml sample of 0.250 M HCN is titrated with 50.0 ml of 0.200 M NaOH. What are the concentrations of HCN and CN in solution after the neutralization reaction occurs?

Explanation / Answer

1)

molecular weight of KOh=56.10

moles of KOh in 10.6 gm=10.6/56.10= 0.188948307 moles in 125 ml

so in 10 ml of soln, moles will be= 0.188948307*10/125= 0.0151158646 moles

0.0151158646 moles is present in 250 ml so molarity =0.0151158646/0.250= 0.0604634584 M

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