what is the hydrogen ion concentration of an aqueous solution with a pOH of 9.262?
I got 5.47X 10 ^-10
is this correct?
Thanks Andy
Yes, that is correct.
[H+] = antilog(9.262)
= 10^(-9.262)
= 5.47x10^-10
I'm not ure that that is right...
[H+] = -log(pH)
however, we are given a pOH, not a pH.
pH + pOH = 14
pH = 14 - pOH
so 14 - 9.262 = pH
pH = 4.738
[H+} = 10^-ph = 10^-4.738
[H+] = 1.8281 x 10^-5 M
To find the hydrogen ion concentration (H+) of an aqueous solution given the pOH, we can use the following formula:
pH + pOH = 14
Given that the pOH is 9.262, we can subtract it from 14 to find the pH:
pH = 14 - pOH
pH = 14 - 9.262
pH = 4.738
The pH represents the negative logarithm of the hydrogen ion concentration, so we can calculate the hydrogen ion concentration by taking the antilog of the pH value:
[H+] = 10^(-pH)
[H+] = 10^(-4.738)
[H+] = 4.47 × 10^(-5) or 4.47E-5
Therefore, the correct hydrogen ion concentration of the aqueous solution is 4.47 × 10^(-5) or 4.47E-5. It appears that your answer of 5.47 × 10^(-10) is not correct.
To find the hydrogen ion concentration (or pH) of an aqueous solution, we can use the relationship between pH, pOH, and the concentration of hydrogen ions (H+). The equation is as follows:
pH + pOH = 14
Given that the pOH of the solution is 9.262, we can find the pH by subtracting it from 14:
pH = 14 - pOH
pH = 14 - 9.262
pH = 4.738
Therefore, the pH of the solution is approximately 4.738.
Now, to find the hydrogen ion concentration (H+), we can use the equation:
pH = -log[H+]
Rearranging the equation to find the concentration:
[H+] = 10^(-pH)
Plugging in the value we found for pH:
[H+] = 10^(-4.738)
[H+] = 4.668 × 10^(-5)
So, the hydrogen ion concentration of the solution is approximately 4.668 × 10^(-5) or 5.47 × 10^(-10). Thus, your answer of 5.47 × 10^(-10) is correct.
Remember to always keep track of the correct number of significant figures in your calculations when reporting the final answer.