starting from the relationship kw = 5.47 * 10^-14 at 323 k, calculate the ph of
ID: 945731 • Letter: S
Question
starting from the relationship kw = 5.47 * 10^-14 at 323 k, calculate the ph of water at this temperature. explain any assumptions you make in arriving at your answer. starting from the relationship kw = 5.47 * 10^-14 at 323 k, calculate the ph of water at this temperature. explain any assumptions you make in arriving at your answer. starting from the relationship kw = 5.47 * 10^-14 at 323 k, calculate the ph of water at this temperature. explain any assumptions you make in arriving at your answer.Explanation / Answer
Kw = 1*10-14 mol2*dm-6 at 25 Celsius degrees
Kw is also called the ionic product of water and it varies with temperature
Kw = [H+]*[OH-]
But [H+]=[OH-] => Kw = [H+]2
Replace the value of Kw:
5.47*10-14 = [H+]2
Take square root both sides:
[H+] = sqrt 5.47*10-14
[H+] = 2.33*10-7
pH = -lg 2.33*10-7
pH = lg (1/2.33) - lg 10-7
pH = 7 - 0.36
pH = 6.64
Therefore pH of water at this temperature is 6.64
Since pH is smaller than 7, then an acid solution is dissolved in water.
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