i have difficulty in this question. any help would be appreciated.
1) What pH corresponds to each of the following: (a) (H+)=0.35M,(b)(OH^-)= 6.0*10^-6M. (d)(H+)= 2.5*10^-8M?
To determine the pH corresponding to each of these values, we can use the equation:
pH = -log[H+]
Where [H+] is the concentration of the hydrogen ions (H+) in the solution. Let's calculate the pH for each scenario:
a) (H+) = 0.35 M
pH = -log(0.35)
pH ≈ -(-0.455)
pH ≈ 0.455
Therefore, the pH corresponding to (H+) = 0.35 M is approximately 0.455.
b) (OH-) = 6.0 × 10^-6 M
To determine the pH using the concentration of hydroxide ions (OH-), we need to use the equation:
pOH = -log[OH-]
Since pH + pOH = 14, we can find the pH by subtracting the pOH from 14:
pOH = -log(6.0 × 10^-6)
pOH ≈ -(-5.222)
pOH ≈ 5.222
pH = 14 - pOH
pH ≈ 14 - 5.222
pH ≈ 8.778
Therefore, the pH corresponding to (OH-) = 6.0 × 10^-6 M is approximately 8.778.
c) (H+) = 2.5 × 10^-8 M
pH = -log(2.5 × 10^-8)
pH ≈ -(-7.602)
pH ≈ 7.602
Therefore, the pH corresponding to (H+) = 2.5 × 10^-8 M is approximately 7.602.
I hope this helps! Let me know if you have any further questions.
To find the pH corresponding to different concentrations of H+ or OH-, you need to know the relationship between pH and the concentration of hydrogen ions (H+), which is measured on a logarithmic scale.
The pH scale ranges from 0 to 14. A pH of 7 is considered neutral, pH values below 7 are acidic, and pH values above 7 are basic or alkaline. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration.
To solve the given question, we will use the pH formula:
pH = -log[H+]
(a) H+ = 0.35M
To find the pH corresponding to this concentration, substitute the value of [H+] into the formula:
pH = -log(0.35) = 0.455
So, the pH corresponding to (a) is approximately 0.455.
(b) OH- = 6.0 * 10^-6M
Since we are given the concentration of OH- rather than H+, we need to calculate the concentration of H+ using the equilibrium constant expression for water:
Kw = [H+][OH-] = 1.0 x 10^-14
Rearranging the equation, we have:
[H+] = Kw / [OH-] = (1.0 x 10^-14) / (6.0 x 10^-6)
[H+] = 1.67 x 10^-9M
Now, substitute this value of [H+] into the pH formula:
pH = -log(1.67 x 10^-9) ≈ 8.78
Thus, the pH corresponding to (b) is approximately 8.78.
(d) H+ = 2.5 * 10^-8M
Using the pH formula:
pH = -log(2.5 x 10^-8) ≈ 7.6
Therefore, the pH corresponding to (d) is approximately 7.6.