Given the following functions, answer the questions:
Temperature (°C) Kw
0 1.1410−15
25 1.0010−14
35 2.0910−14
40. 2.9210−14
50. 5.4710−14
Is the autoionization of water exothermic or endothermic?
(b) What is the pH of pure water at 35°C?
(c) From a plot of ln(Kw) versus 1/T (using the Kelvin scale), estimate Kw at 38°C. (Give your answer to three significant figures.)
(d) What is the pH of a neutral solution at 38°C? (Give your answer to three decimal places.)
I was able to get the first part but I do not understand the rest of it
Ok. You have a partg.
b.
Kw = 2.09E-14; therefore,
(H^+) = sqrt(2.09E-14)
(H^+) = 1.44E-7 and pH = 6.84
c. You will need to plot.
d. Read the plot you made in c and follow the example I did in part b.I would guess Kw to be about 2.6 or 2.7E-14 at 38C.
I'm sorry but I am still confused.
To answer the remaining questions, we can use the information about the autoionization constant of water, Kw, at different temperatures. Kw represents the equilibrium constant for the reaction when water molecules dissociate into hydronium (H3O+) and hydroxide (OH-) ions.
(b) To find the pH of pure water at 35°C, we need to calculate the concentration of hydronium ions ([H3O+]). Since pure water is neutral, the concentration of hydronium ions is equal to the concentration of hydroxide ions ([OH-]).
We can use the equation Kw = [H3O+][OH-] to find the concentration of hydronium ions. From the given data, at 35°C, Kw = 2.0910^-14. Since [H3O+] = [OH-], we can assume x as the concentration of hydronium ions. Thus,
Kw = (x)(x) = x^2
From this equation, we can solve for x:
x^2 = 2.0910^-14
Taking the square root of both sides, we get:
x = sqrt(2.0910^-14) ≈ 1.447×10^-7
Since the pH is defined as -log[H3O+], the pH of pure water at 35°C is approximately -log(1.447×10^-7) = 6.839.
(c) To estimate Kw at 38°C, we can use the given data to create a plot of ln(Kw) versus 1/T (temperature in Kelvin).
Using the temperature values provided, we can convert them to Kelvin by adding 273.15:
Temperature (°C) K (Kelvin)
0 273.15
25 298.15
35 308.15
40. 313.15
50. 323.15
Plotting ln(Kw) on the y-axis and 1/T on the x-axis, we can draw a best-fit line or curve. From this plot, we can estimate the value of ln(Kw) at 1/T = 1/311.15 K^-1 (corresponding to 38°C in Kelvin).
Using the calculated value from the plot, we can find Kw by taking the exponential of ln(Kw):
Kw = e^(ln(Kw estimation))
Round the value to three significant figures to provide the estimated value of Kw at 38°C.
(d) The pH of a neutral solution is 7. At 38°C, the pH of a neutral solution is still 7, as it remains neutral regardless of the temperature.
Remember that for (c), you will need to create a plot and estimate the value of Kw at 38°C based on the plot.