A 0.600 g sample of potassium carbonate, K₂CO3 (138.2055 g/mol) is dissolved in 20.00 mL of 0.1700 M hydrochloric acid, HCI. The resulting solution was then titrated with 0.1048 M sodium hydroxide, NaOH and required 24.16 mL to reach end point. Calculate the percentage (w/w) of potassium carbonate in the sample. K₂CO3(aq) + 2HCl(aq) NaOH(aq) + HCl(aq) 2KCl(aq) + H₂O(1) + CO2(g) NaCl(aq) + H₂O(1)
A 0.600 g sample of potassium carbonate, K₂CO3 (138.2055 g/mol) is dissolved in 20.00 mL of 0.1700 M hydrochloric acid, HCI. The resulting solution was then titrated with 0.1048 M sodium hydroxide, NaOH and required 24.16 mL to reach end point. Calculate the percentage (w/w) of potassium carbonate in the sample. K₂CO3(aq) + 2HCl(aq) NaOH(aq) + HCl(aq) 2KCl(aq) + H₂O(1) + CO2(g) NaCl(aq) + H₂O(1)
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Problem 2.1ACP: Assume that a sample of hard water contains 50. mg/L of Mg2+ and 150 mg/L of Ca2+, with HCO3 as the...
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![A 0.600 g sample of potassium carbonate, K₂CO3 (138.2055 g/mol) is dissolved in
20.00 mL of 0.1700 M hydrochloric acid, HCI. The resulting solution was then titrated
with 0.1048 M sodium hydroxide, NaOH and required 24.16 mL to reach end point.
Calculate the percentage (w/w) of potassium carbonate in the sample.
K₂CO3(aq) + 2HCl(aq)
NaOH(aq) + HCl(aq)
2KCl(aq) + H₂O(1) + CO2(g)
NaCl(aq) + H₂O(1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4d8f1d29-7285-42c7-b9b7-db17e2c86bcc%2F6d430bc4-b2d5-4d70-8e86-0c7fbb0a5114%2Fjexnxeh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 0.600 g sample of potassium carbonate, K₂CO3 (138.2055 g/mol) is dissolved in
20.00 mL of 0.1700 M hydrochloric acid, HCI. The resulting solution was then titrated
with 0.1048 M sodium hydroxide, NaOH and required 24.16 mL to reach end point.
Calculate the percentage (w/w) of potassium carbonate in the sample.
K₂CO3(aq) + 2HCl(aq)
NaOH(aq) + HCl(aq)
2KCl(aq) + H₂O(1) + CO2(g)
NaCl(aq) + H₂O(1)
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