Question

A student dissolves 5.00 g calcium hydroxide to make 100.0 mL of solution. If 63.5 mL...

A student dissolves 5.00 g calcium hydroxide to make 100.0 mL of solution. If 63.5 mL of this solution is required to neutralize 150.0 mL of a hydrofluoric acid solution, what is the concentration of the original HF solution? (The molar mass of Ca(OH)2 is 74.10 g/mol.)

I need help with this asap please.

Homework Answers

Answer #1

5.00 g is dissolved in 100.0 mL

So,

63.5 mL will contain, 5*63.5/100 = 3.175 g

This should be neutralised

we have the Balanced chemical equation as:

Ca(OH)2 + 2 HF ---> CaF2 + 2 H2O

Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

mass of Ca(OH)2 = 3.175 g

we have below equation to be used:

number of mol of Ca(OH)2,

n = mass of Ca(OH)2/molar mass of Ca(OH)2

=(3.175 g)/(74.096 g/mol)

= 4.285*10^-2 mol

From balanced chemical reaction, we see that

when 1 mol of Ca(OH)2 reacts, 2 mol of HF reacts

mol of HF reacted = (2/1)* moles of Ca(OH)2

= (2/1)*4.285*10^-2

= 8.57*10^-2 mol

This is number of moles of HF

volume , V = 150.0 mL

= 0.15 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 8.57*10^-2/0.15

= 0.5713 M

Answer: 0.571 M

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
A student dissolves 153 milligrams of a diprotic acid in water and titrates the solution with...
A student dissolves 153 milligrams of a diprotic acid in water and titrates the solution with a 0.150 M solution of sodium hydroxide. It takes 27.15 mL of sodium hydroxide to neutralize the acid. What is the molar mass (in g/mol) of the acid?
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.988 g...
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.988 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 18.7 mL of calcium hydroxide to reach the endpoint. A. What is the molarity of the calcium hydroxide solution? This calcium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid. B. If 12.7 mL of the calcium hydroxide solution is required to neutralize 14.9 mL of hydroiodic acid,...
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.905 g...
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.905 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 31.0 mL of calcium hydroxide to reach the endpoint. A. What is the molarity of the calcium hydroxide solution? This calcium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid. B. If 25.0 mL of the calcium hydroxide solution is required to neutralize 19.3 mL of hydrochloric acid,...
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.925 g...
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.925 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 28.3 mL of calcium hydroxide to reach the endpoint. A. What is the molarity of the calcium hydroxide solution? This calcium hydroxide solution is then used to titrate an unknown solution of hydroiodic acid. B. If 13.3 mL of the calcium hydroxide solution is required to neutralize 29.9 mL of hydroiodic acid,...
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-]...
During an experiment, a student adds 0.339 g of calcium metal to 100.0 mL of 2.05...
During an experiment, a student adds 0.339 g of calcium metal to 100.0 mL of 2.05 M HCl. The student observes a temperature increase of 11.0 °C for the solution. Assuming the solution\'s final volume is 100.0 mL, the density is 1.00 g/mL, and the specific heat is 4.184 J/(g·°C), calculate the heat of the reaction, ΔHrxn. Ca(s)+2H+(aq) ---->Ca2+(aq) + H2(g) delta Hrxn= ______kJ/mol
A student added 50.0 mL of a sodium hydroxide solution to 100.0 mL of 0.400 M...
A student added 50.0 mL of a sodium hydroxide solution to 100.0 mL of 0.400 M HCl. The resulting solution was then treated with an excess of aqueous chromium (III) nitrate, resulting in the formation of 2.06 g of precipitate. Determine the concentration of the initial sodium hydroxide solution.
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.921g potassium...
A student is asked to standardize a solution of calcium hydroxide. He weighs out 0.921g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 17.3 mL of calcium hydroxide to reach the endpoint. A. What is the molarity of thecalcium hydroxide solution? M=? his calcium hydroxidesolution is then used to titrate an unknown solution ofhydrobromic acid. B. If 19.5 mL of the calcium hydroxide solution is required to neutralize 14.2 mL ofhydrobromic acid, what is the molarity...
When 10.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is...
When 10.0 g of calcium metal is reacted with water, 5.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction? (6 pts) Ca (s) + 2 H2O (l) → Ca(OH)2 (aq) + H2 (g)
A 1.224-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of...
A 1.224-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 55.0 mL of this solution was titrated with 0.08096-M NaOH. The pH after the addition of 12.88 mL of base was 7.00, and the equivalence point was reached with the addition of 41.81 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What is the molar mass of the...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT