A 720 mL sample of water was cooled from 50. 0 °C to 10. 0 °C. How much heat was lost?.
1 answer:
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Answer:
See explanation below
Explanation:
I'm assuming this is a table you need to complete, so, you are not putting this in order, but I already found it in another place. let's do a little summary of the expressions we need to use in order to complete the chart.
To calculate pH we need the following expression:
<em>pH = -log[H3O+] (1)</em>
From this expression we can solve for [H3O+] in case we need it:
<em>[H3O+] = 10^(-pH) (2)</em>
When we have [OH-] we calculate the pOH and from there, the pH:
<em>pOH = -log[OH-] (3)</em>
and [OH-]:
<em>[OH-] = 10^(-pOH) (4)</em>
Finally to get the pH from pOH:
14 = pH + pOH
<em>pH = 14 - pOH (5)</em>
With these 5 expressions we can complete the chart. In picture 1, you have the actual chart.
To know if it's acidic or basic, that depends on the value of pH.
If pH <7 it's acidic
If pH >7 it's basic
If pH = 7 it's neutral
<u><em>First case:</em></u>
[H3O+] = 3.5x10^-3
pH = -log(3.5x10^-3) = 2.46 (acidic)
pOH = 14 - 2.46 = 11.54
[OH-] = 10^(-11.54) = 2.88x10^-12 M
<u><em>Second case:</em></u>
pOH = -log(3.8x10^-7) = 6.42
pH = 14 - 6.42 = 7.58 (it's basic)
[H3O+] = 10^(-7.58) = 2.63x10^-8 M
<u><em>Third case:</em></u>
pH = -log(1.8x10^-9) = 8.74 (it's basic)
pOH = 14 - 8.74 = 5.26
[OH-] = 10^(-5.26) = 5.5x10^-6 M
<u><em>Fourth case:</em></u>
[H3O+] = 10^(-7.15) =7.08x10^-8 M (Basic)
pOH = 14 - 7.15 = 6.85
[OH-] = 10^(-6.85) = 1.41x10^-7 M
Hope this can help you
