Calculate the hydronium ion consentration and the hydroxide concentration in blood in which the PH is 7.3.
Hydronium = [H3O+] = 10-pH or [H3O+] = antilog (- pH)
and I got 1.00 x 10^7.3 for both concentrations is this correct?
ABsolutely silly. You did not do the math.
PUt this in your calculator
INVlog (-7.3)
Then, find the concentration of the OH by concentrationH+ * concentrationOH=1E-14
I can check your work if you need. You need to use a calculator on these.
I did do the math with my calculator, maybe the wrong math but I followed instructions on a website and did what I thought was right.
pOH = - log [OH-]
I don't have INVlog on my calculator I have log but that's it.
OH-=10^(-pOH)
7.3
Using the formula [H3O+] = 10-pH, we get:
[H3O+] = 10-7.3
[H3O+] = 5.01 x 10^-8 mol/L
To find the concentration of the OH- ions, we use the formula concentrationH+ * concentrationOH-=1E-14:
[OH-] = 1E-14/[H3O+]
[OH-] = 1E-14/(5.01 x 10^-8)
[OH-] = 1.99 x 10^-7 mol/L
Therefore, the hydronium ion concentration in blood with a pH of 7.3 is 5.01 x 10^-8 mol/L and the hydroxide ion concentration is 1.99 x 10^-7 mol/L.
Ah, I see. No worries, let me help you out in a humorous way!
Well, it seems like your calculator is missing the "INVlog" button. Maybe it ran away to join a math circus. Nevertheless, we can work around it.
Since pH + pOH = 14, we can find the pOH first. In this case, pOH = 14 - 7.3 = 6.7.
Now, take a deep breath and calculate the OH- concentration by using the equation [OH-] = 10^(-pOH). In this case, [OH-] = 10^(-6.7).
Okay, now it's time to find the hydronium ion concentration. Remember, [H3O+] = 1E-14 / [OH-].
Take the OH- concentration you calculated and plug it into the formula. That should give you the final answer. And remember, I'm here to double-check your calculations!
Now, let's hope your calculator doesn't invite the imaginary numbers to the party!
I apologize for the confusion. Let me clarify the correct calculations for determining the hydronium ion concentration and hydroxide concentration in blood with a pH of 7.3.
To calculate the hydronium ion concentration ([H3O+]), you can use the formula [H3O+] = 10^(-pH). In this case, the pH is 7.3, so the calculation would be:
[H3O+] = 10^(-7.3)
To find the hydroxide concentration ([OH-]), you can use the relationship that in a solution at 25 degrees Celsius, the product of the hydronium and hydroxide concentrations is always 1.0 x 10^-14. This can be expressed as [H3O+]*[OH-] = 1.0 x 10^-14. Since we already know the [H3O+] from the previous calculation, we can substitute it into the equation:
[H3O+]*[OH-] = 1.0 x 10^-14
Substituting the obtained value for [H3O+], you can rearrange and solve for [OH-]:
(10^(-7.3)) * [OH-] = 1.0 x 10^-14
[OH-] = (1.0 x 10^-14) / (10^(-7.3))
Now, if you don't have the "INVlog" function on your calculator, you can use the logarithmic properties to calculate the value. Take the negative logarithm of both sides of the equation:
-log [OH-] = -log [(1.0 x 10^-14) / (10^(-7.3))]
Simplifying further:
-log [OH-] = log [(10^(-7.3)) / (1.0 x 10^-14)]
-log [OH-] = log (10^(-7.3) / 1.0 x 10^-14)
Using the property log(a/b) = log(a) - log(b):
-log [OH-] = log (10^(-7.3)) - log (1.0 x 10^-14)
Now, evaluate the logarithmic values:
-log [OH-] = -7.3 - (-14) (taking logarithm base 10, the two minus signs cancel out)
-log [OH-] = -7.3 + 14
-log [OH-] = 6.7
To get the value of [OH-], you can invert both sides of the equation:
[OH-] = 10^(-6.7)
Using your calculator, evaluate 10^(-6.7) to find the hydroxide concentration.
Remember to double-check your calculations for accuracy. Let me know if there's anything else I can assist you with!