Knowee
Questions
Features
Study Tools

20 mL of calcium hydroxide was consumed when it was reacted with   10 mL of unknown solution of H2SO4. Also   20 mL standard solution of     0.5 M HCl containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of   35.5 mL whereas the burette showed   25.5 mL initially. The concentration of H2SO4 is

Question

20 mL of calcium hydroxide was consumed when it was reacted with   10 mL of unknown solution of H2SO4. Also   20 mL standard solution of     0.5 M HCl containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of   35.5 mL whereas the burette showed   25.5 mL initially. The concentration of H2SO4 is

🧐 Not the exact question you are looking for?Go ask a question

Solution

To find the concentration of the H2SO4 solution, we first need to find the concentration of the calcium hydroxide solution. We can do this using the data from the titration with the HCl solution.

  1. Calculate the volume of calcium hydroxide used in the titration with HCl:

    Volume = Final burette reading - Initial burette reading Volume = 35.5 mL - 25.5 mL = 10 mL

  2. Convert the volume to liters:

    Volume = 10 mL = 0.01 L

  3. Calculate the moles of HCl:

    Moles = Molarity x Volume Moles = 0.5 M x 0.01 L = 0.005 moles

  4. Since the reaction between HCl and calcium hydroxide is a 1:1 ratio, the moles of calcium hydroxide is also 0.005 moles.

  5. Calculate the concentration of calcium hydroxide:

    Concentration = Moles / Volume Concentration = 0.005 moles / 0.02 L = 0.25 M

Now we can find the concentration of the H2SO4 solution.

  1. Convert the volume of calcium hydroxide used in the reaction with H2SO4 to liters:

    Volume = 20 mL = 0.02 L

  2. Calculate the moles of calcium hydroxide:

    Moles = Concentration x Volume Moles = 0.25 M x 0.02 L = 0.005 moles

  3. Since the reaction between H2SO4 and calcium hydroxide is a 1:1 ratio, the moles of H2SO4 is also 0.005 moles.

  4. Convert the volume of H2SO4 to liters:

    Volume = 10 mL = 0.01 L

  5. Calculate the concentration of H2SO4:

    Concentration = Moles / Volume Concentration = 0.005 moles / 0.01 L = 0.5 M

So, the concentration of the H2SO4 solution is 0.5 M.

This problem has been solved

Similar Questions

100 mL of a raw water sample required 25 mL of N/50 H2SO4 for neutralisation to phenolphthalein end point. After this methyl orange indicator was added to it and the end point was 28.0 mL. Calculate the alkalinity of water as CaCO3 equvivalents in parts per million.a.Alkalinity due to HCO32– =  60ppm Alkalinity due to CO32– = 220 ppmb.Alkalinity due to OH– = 220 ppm Alkalinity due to CO32– = 60 ppmc.Alkalinity due to HCO32– = 220 ppm Alkalinity due to CO32– = 60 ppmd.Alkalinity due to OH– =60  ppm Alkalinity due to CO32– =220 ppm

100 ml of an alkaline water in presence of phenolphthalein, 50 ml of the N/100 HCl. The resulting mixture in presence of methyl orange indicator consumes 35 ml of HCl. Predict the alkalinities and calculate their amounts in ppm as CaCO3 equivalent.

Calcium hydroxide and phosphoric acid react to form calcium phosphate and water. Ca(OH)2 +  H3PO4 --> Ca3(PO4)2 + H2OHow many grams of H3PO4 are needed to produce 1,460 grams of Ca3(PO4)2?Write your final answer in two decimal places.Use the following mass numbers:Ca - 40 g/molH - 1 g/molO - 16 g/molP - 31 g/mol

How much calcium hydroxide is produced by the complete reaction of 150.0 g CaO with 50.0 g of water?

A water sample is alkaline to both phenolphthalein as well as methyl orange. 200 mL of a this water sample on titration with N/50 HCl required 9.4 mL of the acid to phenolphthalein end point. When a few drops of methyl orange are added to the same solution and titration is further continued, the yellow color of the solution just turned red after addition of 30.4 mL of the acid solution. Calculate the  alkalinity present in the water sample.a.P = 152 ppm., M =  47ppm and HCO3– alkalinity = 58 ppm, CO32– alkalinity = 94 ppmb.P = 1520 ppm., M =  470ppm and HCO3– alkalinity = 940 ppm, CO32– alkalinity = 580 ppmc.P = 47 ppm., M = 152 ppm and HCO3– alkalinity = 58 ppm, CO32– alkalinity = 94 ppmd.P = 470 ppm., M = 1520 ppm and HCO3– alkalinity = 580 ppm, CO32– alkalinity = 940 ppm

1/3

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.