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From: J♥i♥m♥i♥n♥r♥o♥l♥l♥s

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If you ever swam in a pool and your eyes began to sting and turn red, you felt the effects of an incorrect pH level. pH measures

Kobotan [32]

For a better understanding of the explanation provided here, please go through the graphs in the attached files. The first file contains the explanations to the first part of the question and the second file contains the three graphs required to explain the second part (the y intercept part) of the question.

It has been given to us that the pH concentration equation is:

$p(t)=-log_{10}t$

where, The variable t represents the amount of hydronium ions and p(t) gives the resulting pH level.

As can be seen from the first graph, pH is 1 when the amount of hydronium ions, t is 0.1. This is as expected, because, from the concept of logarithms we know that:

$log_{10}(0.1)=log_{10}(\frac{1}{10})=log_{10}(10^{-1})=-1\times log_{10}10=-1\times 1=-1$

Thus, $p(0.1)=-1\times-1=1$

Again, the value of pH, p(t) is 0 when the concentration of the Hydronium ions, t is 1. This, again, is as expected, because, from the concept of logarithms we know that $log(1)=0$ .

If the pool maintenance man uses our graph and needs to find the the pH level if the amount of hydronium ions is raised to 0.50, then to do so, the pool maintenance man must draw a vertical line from t=0.5 and check the x coordinate of the point where this vertical line intersects the p(t) graph. As we can see from the graph this roughly comes out to be the horizontal line y=0.3.

Let us now move on to the second part of the question. For a better understanding of this part please go through the second graph which has been attached. This graph contains all translations on the original graph and the explanation to each one of them is provided below:

Let us start with the last one, -p(t). We know that, since, $p(t)=-log_{10}(t)$ , therefore, $-p(t)=log_{10}(t)$ , simply a vertical "flip" or reflection of the original graph. And because the original graph did not have a y intercept, this graph too will not have a y intercept.

Let us now move on to the first one. Here, $p(t)+1=-log(t)+1$ . This graph will simply move the original graph vertically up by 1 unit and that will not have any bearing on the y intercept and thus, the graph of p(t)+1 will again, not have a y intercept.