Asked by chris
Understanding
and Ion Concentrations
Use Table B in your Student Guide to answer the questions about ion concentrations.
A solution with a
has approximately how many moles of
ions per liter?
How many moles of
would this same solution have per liter?
(Use the decimal form of your answer.)
A different solution with an
concentration of
would have a
.
All Answers 1
Answered by
GPT-5 mini
AI
I don't have your Table B or the missing numbers/ions in the question, so I can't fill the exact blanks. Could you paste the table or the full text with the missing values?
Meanwhile, here is the method and a common example (so you can check or apply it to your specific numbers):
Method:
- For pH problems: [H+] = 10^(−pH). [OH−] = Kw / [H+] with Kw = 1.0 × 10^−14 at 25 °C. To get pH from [H+]: pH = −log10([H+]).
Example (if the blank was “pH = 3” and the ions are H+ and OH−):
- [H+] = 10^(−3) = 0.001 mol/L.
- [OH−] = 1.0×10^(−14) / 0.001 = 1.0×10^(−11) = 0.00000000001 mol/L.
- If a different solution has [H+] = 1.0×10^(−6) mol/L, pH = −log10(1.0×10^(−6)) = 6.
If your problem uses different values or different ions, paste them and I’ll compute the exact decimal answers.
Meanwhile, here is the method and a common example (so you can check or apply it to your specific numbers):
Method:
- For pH problems: [H+] = 10^(−pH). [OH−] = Kw / [H+] with Kw = 1.0 × 10^−14 at 25 °C. To get pH from [H+]: pH = −log10([H+]).
Example (if the blank was “pH = 3” and the ions are H+ and OH−):
- [H+] = 10^(−3) = 0.001 mol/L.
- [OH−] = 1.0×10^(−14) / 0.001 = 1.0×10^(−11) = 0.00000000001 mol/L.
- If a different solution has [H+] = 1.0×10^(−6) mol/L, pH = −log10(1.0×10^(−6)) = 6.
If your problem uses different values or different ions, paste them and I’ll compute the exact decimal answers.
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