. 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.
Chapter7: Solutions And Colloids
Section: Chapter Questions
Problem 7.31E
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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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