The average pH of normal arterial blood is 7.40. At normal body t...

Question

The average pH of normal arterial blood is 7.40. At normal body temperature 137 °C2, Kw = 2.4 * 10-14. Calculate 3H+4, 3OH-4, and pOH for blood at this temperature.

Accepted Answer

Hello everyone. So in this video we're trying to solve for the concentration of H plus ions concentration of R O. H minus ions and the P. O. H. Given the ph of a solution as well as the K. W. So first things first we see that the first piece of information that were given to us in this problem is the ph and that is a value 5.60. We can go ahead and calculate for H plus because if you recall the equation for that is 10 to the negative ph so sorry for that. We see we have 10 to the negative 5.60. So putting that value into my calculator by concentration of our H plus is going to be 2.5119 times to the negative six Mohler. Now rounding that to the appropriate amount of significant figures, that's going to be 2.5 times 10 to the negative six Mohler. So formally this is going to be my final answer but we're still going to use this value here for future calculations to go ahead and eliminate any errors that we might get. So now we can go ahead and solve for K. W. Kw if your call is equal to the concentration of H plus, multiplied by concentration of R O H minus. We see that we have two values which one of which we are given here in the problem and one of which we have just offered right here. So we're going for the concentration of R. O. H. Here, let's go ahead and rearrange this equation. So we're solving for the concentration of R. O. H ions that's going to equal to the K. W. Over the concentration of H Plus. So plugging in those numerical values KW is on top, It's 1.47, 1 times 10 to the -14. And the concentration of R. H plus is going to be this value here. That is 2.5119 times 10 to the negative six Mohler. Putting those values into my calculator, I will get that. The concentration of my hydroxide ions is going to be 5.8561 times 10 to the -9 moller. And of course writing that to the appropriate amount of significant figures, we're going to report this as 5.9 times 10 to the -9 moller. So that's one of my answers. And lastly we're gonna go ahead and solve for R P. O. H. So P. O. H. If you recall the the equation for this is negative log of the concentration of O. H minus again which is what we just saw for. So plugging the value in the negative log of 5.9 times 10 to the -9, that's giving me a P. O. H. Value of 8.23. And this is going to be my last answer for this problem. So again our concentration of H plus is going to be 2.5 times to the negative six moller. Our concentration of our hydroxide ions is 5.9 times 10 to the negative nine moller. And the p o. H. Of this solution is 8.23. Thank you all so much for watching.

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