A solution contains Na2CO3 and NaHCO3. 10 mL of the solution required 2.5 mL of 0.1 M H2SO4 for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 mL of 0.2 M H2SO4 was required. The amount of Na2CO3 and NaHCO3 in 1 litre of the solution is: (a) 5.3g and 4.2g (b) 3.3g and 6.2g (c) 4.2g and 5.3g (d) 6.2g and 3.3g

A solution contains Na2CO3 and NaHCO3. 10 mL of the solution required 2.5 mL of 0.1 M H2SO4 for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 mL of 0.2 M H2SO4 was required. The amount of Na2CO3 and NaHCO3 in 1 litre of the solution is: (a) 5.3g and 4.2g (b) 3.3g and 6.2g (c) 4.2g and 5.3g (d) 6.2g and 3.3g

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter15: Acid–base Equilibria

Section: Chapter Questions

Problem 99AP

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If the normality of HCl is greater by 10% compared to KOH, estimate the normality of the HCl solution when 0.67 gram of (NH4)2SO4, is dissolved in water, treated with excess NaOH and the ammonia released is passed through a solution of 20 ml of HCl. The excess acid solution is titrated using 15.01 ml of KOH solution.
50 mL sample solution containing Na2CO3 and NaOH is titrated with 0.2 M HCl solution. In the presence of phenolphthalein indicator, the acid consumption is 15 mL. Then, the titration is continued by adding bromocresol green indicator, and in the presence of this indicator, the acid consumption is 10 mL. Accordingly, which of the following is the amount of Na2CO3 (in grams) in the sample? (NaOH: 40 g / mol; NazCO3: 106 g / mol)
A solution contains KHCO3, K2CO3 and KOH either alone or in permissible combination. A 50.0 mL aliquot requires 25.0 mL of 0.100 M HCI to reach the phenolphthalein end point. Another 50.0 mL aliquot requires 45.0 mL of the same acid to reach the bromocresol green end point. Which of the following statements BEST describes the composition of the sample solution? O The solution contains more KOH than K2CO3. O The solution contains K2CO3 only. O The solution contains more KHCO3 than K2CO3. O The solution contains a mixture of KHCO3 and KOH. O The solution contains KOH only. O The solution contains more K2CO3 than KOH.
A 25.00-mL sample of vinegar is diluted with deionized to 100.0 mL. Then a 10.00-mL aliquot (portion) of the dilute vinegar solutionis titrated with 0.1055MNaOHsolution using phenolphthalein as theindicator. (a) If 19.60mL of the base solution is required to reach the end-point, calculate the molar concentration of acetic in dilute vinegar solution. (b) What is the molar concentration of acetic acid in the original (undiluted) vinegar? At the end-point, the volume of NaOH consumed was found to be 32.00 mL. The acid-base reaction occurs as follows: HC2H3O2(aq) + NaOH(aq) --> H2O(l) + NaC2H3O2(aq)
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I want the answer for Q2
In the experiment conducted to determine the amount of acetic acid in vinegar, 8.2 mL of NaOH standard solution adjusted to 0.098 M was spent besides the phenolphthaleley indicator. Accordingly, which of the following is the amount of acetic acid in the vinegar sample in grams? (Ma (CH3COOH): 60.05 g / mol)
A 0.5123g sample (molar mass 204.23g/mol) of KHP was dissolved in about 25 mL of distilled water and titrated to the phenolphthalein end point with 28.75 mL of a sodium hydroxide solution. Calculate the molar concentration of the hydroxide solution.
A 50.00 ml of a sample of river water was titrated with 0.0015M of sodium dichromate and required 11.50 ml to reach the endpoint. What is COD in mg of Oxygen /L and in molecules of oxygen/L?
Which is the best solution to buffer at pH = 10.50? (A) A buffer solution containing 0.25 M H2CO3 and 0.45 M Na2CO3; (Ka (H2CO3) = 4.3 x 10–7) (B) A buffer solution containing 0.25 M NaHCO3 and 0.45 M Na2CO3; (Ka (HCO3–) = 5.6 x 10–11) (C) A buffer solution containing 0.12 M NaHCO3 and 0.22 M Na2CO3; (Ka (HCO3–) = 5.6 x 10–11) (D) A buffer solution containing 0.12 M NH4Cl and 2.2 M NH3; (Ka (NH4+) = 5.6 x 10–10)
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