The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14. Show all calculations leading to an answer.
I got this so far :
Kw = 1.219x 10^-14
[H+]^2= 1.219 x 10^-14
[H+]= -/(1.219) x 10^14
[H+] = 1.104 x 10^-7
pH= -log [H+]
pH= -log [1.104 x 10^-7]
pH= (log 1.104 + log 10^-7)
I don't completely understand how to calculate pH from -log[H+]
Look on your calculator, hit the log button and type in 1.104E-7 and hit the enter button or equal button. Then multiply by -1. I get 6.957 which I would round to 6.96.
To calculate the pH from the concentration of hydrogen ions ([H+]), you need to take the negative logarithm (base 10) of the concentration.
In your case, you have [H+] = 1.104 x 10^-7. Here's how to calculate the pH step-by-step:
1. Take the negative logarithm of [H+]:
pH = -log (1.104 x 10^-7)
2. Simplify the expression:
pH = -log (1.104) - log (10^-7)
3. Use the properties of logarithms:
Recall that log (a * b) = log a + log b.
Since you have 10^-7, which is equivalent to (10^(-1)) * (10^(-6)), you can rewrite it as 10^(-1-6).
pH = -log (1.104) - log (10^(-1-6))
Now, using the logarithmic property, you can rewrite it as:
pH = -log (1.104) - (log (10^(-1)) + log (10^(-6)))
4. Continue simplifying:
Since log (10^(-1)) = -1 and log (10^(-6)) = -6, you can further simplify the expression:
pH = -log (1.104) - (-1 + (-6))
pH = -log (1.104) - (-7)
pH = -log (1.104) + 7
5. Use a calculator to find the logarithm of 1.104:
log (1.104) ≈ -0.043
Substituting this back into the expression:
pH ≈ -(-0.043) + 7
pH ≈ 0.043 + 7
pH ≈ 7.043
So, the pH of water at 46⁰C with a Kw of 1.219 x 10^-14 is approximately 7.043.