更新1:
in this solution? Very lost here. Any help would be amazing.
If I mix 8 mL of 0.50 CH3COOH with 2 ml of 0.50 M NaCH3COO, how do I calculate the concentrations of acetic acid and sodium acetate...?
回答 (2)
2016-03-09 2:13 am
Final volume = 8 + 2 = 10 mL
[CH₃COOH] = 0.5 × (8/10) = 0.4 M
[CH₃COONa] = 0.5 × (2/10) = 0.1 M
[CH₃COOH] = 0.5 × (8/10) = 0.4 M
[CH₃COONa] = 0.5 × (2/10) = 0.1 M
2016-03-09 4:02 am
8 mL x 0.50 M = 4 mmol of acetic acid
2 mL x 0.50 M = 1 mmol of sodium acetate
Total volume = 10 mL
acetic acid concentration = 4 mmol / 10 mL = 0.4 M
spdoium acetate concentration = 1 mmol / 10 mL = 0.1 M
2 mL x 0.50 M = 1 mmol of sodium acetate
Total volume = 10 mL
acetic acid concentration = 4 mmol / 10 mL = 0.4 M
spdoium acetate concentration = 1 mmol / 10 mL = 0.1 M
收錄日期: 2021-05-01 13:05:10
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