In an experiment similar to the one described in this procedure, a saturated solution of Mg(OH)2 was titrated with 0.0500 M HCl

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In an experiment similar to the one described in this procedure, a saturated solution of Mg(OH)2 was titrated with 0.0500 M HCl

solution. The equivalence point was reached when 0.330 mL of the acid had been added to 50.00 mL of the saturated Mg(OH)2.

Chemistry

2 answers:

Karo-lina-s [1.5K] · Answer 1 · 2021-12-13T13:45:24+03:00

The question is incomplete. The complete question is:

In an experiment similar to the one described in this procedure, a saturated solution of Mg(OH)2 was titrated with 0.0500 M HCI solution. The equivalence point was reached when 0.330 mL of the acid had been added to 50.00 mL of the saturated Mg(OH)2 1. How many moles of H+ were added? How many moles of OH were present in the initial solution of magnesium hydroxide? 2. What are the concentrations of OH and Mg2 in the original 50.00 mL solution? 3 Write the equation for dissolving Mg(OH)2 in water 4. Write the corresponding Ksp expression. 5. What is the value of Ksp found from the experimental data above?

Answer:

Ksp=5.4×10-8

Explanation:

We must find find the concentration of base from the information provided. Then we can now find the concentration of hydrogen ions in the acid, concentration of magnesium ions and hydroxide ions in the base and finally the solubility product of the magnesium hydroxide solution as required. See image attached for more details of the solution to the problem.

Kamila [148] · Answer 2 · 2021-12-13T15:47:37+03:00

Answer:

1/ 1.65*10^(-5) mole; 2/ 1.65*10^(-5) mole; 3/ CM of OH- = 3.3*10^(-4) M, CM of Mg2+ = 1.65 * 10^(-4) M

Explanation:

Mg(OH)2 + 2HCl ⇒ MgCl2 + 2H2O

x (mole)

1/ How many moles of H+ were added?

Molarity = mole/ volume(L)
n = CM*V = 0.05 * 0.330 * 10^(-)3 = 1.65 * 10^(-5) mole

2/ How many moles of OH- were present in the initial solution of magnesium hydroxide?

Moles of OH- are equal to moles of H+ needed for the titration. Thus, there are 1.65 * 10^(-5) mole.

3/ What are the concentrations of OH- and Mg2+ in the original 50.00 mL solution?

CM of OH- = mole/volume = 1.65 * 10^(-5)/ (50 mL * 10^(-3)) = 3.3 * 10^(-4) mole/L
CM of Mg2+ = 1/2 * CM of OH- = 1/2 * 3.3 * 10^(-4) = 1.65 * 10^(-4) mole/L