. The total cation content of natural water is often determined by exchanging the cations for hydrogen ions on an ion exchange resin. The cations are adsorbed to the resin which releases H+ (1:1 exchange) which can then be titrated with a base. A 25.0 mL sample of a natural water is diluted to 100 mL with distilled water and 2.0 g of a cation-exchange resin was added. After stirring, the mixture was filtered and the solid remaining on the filter paper was washed with three 15.0 mL portions of distilled water. The filtrate and washing required 15.3 mL of 0.0202 M NaOH to reach the equivalence pt. Calculate the number of moles of cation present in 1.00 L of sample.
. The total cation content of natural water is often determined by exchanging the cations for hydrogen ions on an ion exchange resin. The cations are adsorbed to the resin which releases H+ (1:1 exchange) which can then be titrated with a base. A 25.0 mL sample of a natural water is diluted to 100 mL with distilled water and 2.0 g of a cation-exchange resin was added. After stirring, the mixture was filtered and the solid remaining on the filter paper was washed with three 15.0 mL portions of distilled water. The filtrate and washing required 15.3 mL of 0.0202 M NaOH to reach the equivalence pt. Calculate the number of moles of cation present in 1.00 L of sample.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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. The total cation content of natural water is often determined by exchanging the cations for hydrogen ions on an ion exchange resin. The cations are adsorbed to the resin which releases H+ (1:1 exchange) which can then be titrated with a base. A 25.0 mL sample of a natural water is diluted to 100 mL with distilled water and 2.0 g of a cation-exchange resin was added. After stirring, the mixture was filtered and the solid remaining on the filter paper was washed with three 15.0 mL portions of distilled water. The filtrate and washing required 15.3 mL of 0.0202 M NaOH to reach the equivalence pt.
Calculate the number of moles of cation present in 1.00 L of sample.
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