Home
Class 12
CHEMISTRY
When 100 ml of a tube-well water is titr...

When 100 ml of a tube-well water is titrated using methyl orange as indicator, it requires 15 ml of 0.01 (N) HCl. Estimate the hardness. Of that sample of water.

Text Solution

Verified by Experts

15 mL 0.01 (N) HCl `equiv` 1 mL 0.15(N) HCl
1000 mL 1 (N) HCl `equiv (100)/(2)g " CaCO"_(3)`
`therefore` 1 mL 0.15 (N) HCl = 50 `xx (1)/(1000) xx 0.15 g " of" "CaCO"_(3)`
`" " 7.5 xx 10^(-3)g " of" " CaCO"_(3)`
Therefore, in 100mL or 100g of that sample of water contains some hardness producing substance which is equivalent to `7.5xx 10^(-3)`g of `"CaCO"_(3)`.
`10^(6)`g of water contains the hardness producing substance
equivalent to `(7.5 xx 10^(-3) xx 10^(6))/(100) = 75`g of `"CaCO"_(3)`
Promotional Banner

Topper's Solved these Questions

  • HYDROGEN

    CHHAYA PUBLICATION|Exercise PRACTICE SET -9|5 Videos

  • HYDROGEN

    CHHAYA PUBLICATION|Exercise SHORT-TYPE QUESTIONS|34 Videos

  • HYDROCARBONS

    CHHAYA PUBLICATION|Exercise PRACTICE SET 13|16 Videos

  • MODEL QUESTION PAPER

    CHHAYA PUBLICATION|Exercise Answer the following question (set-3)|130 Videos

Similar Questions

Explore conceptually related problems

10 mL of 0.01 (N) HCl is required for titrating 100 mL of a sample of cold water using methyl orange as indicator. Determine the temporary hardness of that sample of water.

In 100 ml sample of hard water, 100 ml of (N/50) Na_2CO_3 was added and the mixture was boiled and filtered The filtrate was neutralized with 60 ml of (N/50) HCI. Calculate the permanent hardness of water, (sp gr. of hard water = 1)

Can methyl orange indicator be used in the titration of 0.1 M acetic acid with 0.1 (M) NaOH solution? Explain.

This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements. Statement - I : In the titrations of Na_2CO_3 with HCI using methyl orange indicator, the volume required at the equivalence point is twice that of acid required using phenolphthalein indicator. Statement - II: Two moles of HCI are required for complete neutralization of one mole of Na_2CO_3 .

2g of metal carbonate is neutralised completely by 100 mL of 0.1 (N) HCl. The equivalent weight of metal carbon ate is-

1.37g of a mixture of Na_(2)CO_(3) and NaHCO_(3) was dissolved in water. The solution was titrated with 0.5(M) HCl solution first with phenolphthalein followed by methyl orange indicator. The first and second titrations required 10 mL and 30mL of acid solution respectively. calculate the amounts (in g) of Na_(2)CO_(3) and NaHCO_(3) present in the mixture.

Calculate the pH of a solution which contains 100 ml of 0.1 M HCl and 9.9 ml 1 M NaOH

100 ml of a sample of water required 1.96 mg of potassium dichromate in the presence of 50% H_2SO_4 for the oxidation in the presence organic matter in it. Calculate the chemical oxygen demand.

0.804 g sample of iron was dissolved in acid , Iron was reduced to +2 state and it required 47.2 mL of 0.112 N KMnO_4 solution for titration. Calculate the percentage of iron and Fe_3O_4 in the ore.