Question

I titrated a solution of 25.0 mL of a 0.050 M NaOH solution containing Ca(OH)2 using...

I titrated a solution of 25.0 mL of a 0.050 M NaOH solution containing Ca(OH)2 using 27.71 mL of standardized 0.065 M HCl.

1.) use the HCl volume and concentration to calculate [OH- ] in the saturated solution of Ca(OH)2 in NaOH solution.

2.) Using the known concentration of NaOH solution in Part B, calculate [OH- ] due to the NaOH alone.

3.) Calculate [OH- ] due to dissolved Ca(OH)2.

4.) Calculate [Ca^2+ ].

5.) Calculate Ksp for Ca(OH)2 in the saturated solution of Ca(OH)2 in NaOH.

6.) Calculate the solubility (in g/L) for Ca(OH)2 in the NaOH solution.

Homework Answers

Answer #1

1. The moles of OH- in the saturated solution of Ca(OH)2 in NaOH is equal to the moles of HCl used.

Moles of OH- = MxV = 0.065 M x 0.02771 L = 0.001801 mol

Hence concentration of OH-, [OH-] = 0.001801 mol / 0.025L = 0.0720 M (answer)

2. [OH-] due to NaOH alone = MxV = 0.050 M x 0.025 L / 0.025 L = 0.050 M (answer)

3. [OH-] due to dissolved Ca(OH)2 = 0.0720 - 0.050 = 0.022 M (answer)

4. Ca(OH)2 <---- > Ca2+(aq) + 2 OH-(aq)

1 mol-------------- 1 mol, ----- 2 mol

[Ca2+(aq)] =  0.022 M / 2 = 0.011 M (answer)

5. Ksp = [Ca2+(aq)] x [OH-(aq)]2  = (0.011 M) x(0.022M)2 = 5.32x10-6 (answer)

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
0.050 M HCl Volume at equivalence point = 10.37 mL Volume of Ca(OH)2 used in the...
0.050 M HCl Volume at equivalence point = 10.37 mL Volume of Ca(OH)2 used in the titraiton = 15 mL a) What happens at equivalence point? b) Calculate the [OH-]. c) Calculate the [Ca+2]. d) Calculate the Ksp for Ca(OH)2. e) What is the concentration (in molarity) of the saturated Ca(OH)2?
Question Molar Solubility and solubility Product of Calcium Hydroxide: 1. Volume of Saturated Ca(OH)2 solution (mL)...
Question Molar Solubility and solubility Product of Calcium Hydroxide: 1. Volume of Saturated Ca(OH)2 solution (mL) = 25.0 mL 2. Concentration of Standardized HCL solution (mol/L) = 0.05 mol/L 3. Buret reading, initial (mL) = 50.0 mL 4. Buret reading, final (mL) = 32.5 mL 5. Volume of HCL added (mL) = 17.5 mL 6. Moles of HCL added (mol) = (concentration of HCl)(Volume of HCl) = (0.05)(19.1) = 0.875 mol. 7.Moles of OH- in saturated solution (mol)=8.75x10^-4 8.the [OH-]...
A group of students conducted a titration of 25.00-mL saturated Ca(OH)2 solution with 0.0480M HCl. The...
A group of students conducted a titration of 25.00-mL saturated Ca(OH)2 solution with 0.0480M HCl. The students found that 10.10-mL of acid was required to reach the equivalence point. Calculate the molar concentration of OH- and Ca2+, the molar solubility, and Ksp of the analyte. Show detailed calculation for each. [OH-] in the analyte: [Ca2+] in the analyte: molar solubility of Ca(OH)2: Ksp of Ca(OH)2:
13. A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the...
13. A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)2 as it could hold. A 97.7-mL sample of this solution was withdrawn and titrated with 0.0687 M HBr. It required 57.8 mL of the acid solution for neutralization. (a) What was the molarity of the Ca(OH)2 solution? _____ M (b) What is the solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100...
0.02500 M EDTA solution were used to titrate 10.00 mL saturated Ca(OH)2 solution. EDTA=C10H14N2Na2O8 The average...
0.02500 M EDTA solution were used to titrate 10.00 mL saturated Ca(OH)2 solution. EDTA=C10H14N2Na2O8 The average volume of EDTA solution needed was 11.15 mL to titrate a 10.00 mL saturated solution of Ca(OH)2. milimoles of EDTA= 0.02500(M) *11.15ml = 0.2875milimoles Milimoles of Ca2+= strength * volume = 2.845milimoles ____________________________________________________________ 1.) How many grams of Ca(OH)2 can be dissolved in 1-L solution? 2.) Ksp
A saturated solution of the Ca(Oh)2 is pepared and allowed to equilibrate at 25oC. A 25...
A saturated solution of the Ca(Oh)2 is pepared and allowed to equilibrate at 25oC. A 25 mL sample of the solution is titrated with 0.0050M HCl solution. 1.4mL of the HCl solution are required to fully react with the sample. a) Write a balanced chemical equation of the salt with HCl. b)Calculate the molarity of the salt solution from the titration data. c) Calculate the concentration of the cation and the anion in the salt solution. d)Show the calculation for...
This question is for a laboratory experiment titled "Determination of the Solubility Product Constant of Calcium...
This question is for a laboratory experiment titled "Determination of the Solubility Product Constant of Calcium Hydroxide" where a saturated solution of Ca(OH)2 in 0.02523M NaOH was titrated with HCl. Part 2 Data - Saturated Solution of Ca(OH)2 in 0.02523M NaOH: Volume of Ca(OH)2/NaOH aliquot: 25.00mL Concentration of standard HCl: 0.1342M Indicator Used: Bromothymol Blue Average volume of HCl to reach end point: 10.26mL Ksp from Part 1 of lab: 3.54x10-5 a. Calculate the TOTAL [OH-] in the saturated solution...
A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution...
A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)2 as it could hold. A 50.7-mL sample of this solution was withdrawn and titrated with 0.0621 M HBr. It required 66.7 mL of the acid solution for neutralization. (a) What was the molarity of the Ca(OH)2 solution? ________ M (b) What is the solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100 mL...
A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution...
A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)2 as it could hold. A 93.5-mL sample of this solution was withdrawn and titrated with 0.0700 M HBr. It required 71.5 mL of the acid solution for neutralization. (a) What was the molarity of the Ca(OH)2 solution? ______________ M (b) What is the solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100 mL...
A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution...
A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)2 as it could hold. A 58.6-mL sample of this solution was withdrawn and titrated with 0.0872 M HBr. It required 79.0 mL of the acid solution for neutralization. (a) What was the molarity of the Ca(OH)2 solution? (b)What is the solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100 mL of solution?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT