PLEASE HELP!!! If you ever swam in a pool and your eyes began to sting and turn red, you felt the effects on an incorrect pH level. pH measures the concentration of hydronium ions and can be modeled by the function p(t) = −log10t. The variable t represents the amount of hydronium ions; p(t) gives the resulting pH level. Water at 25 degrees Celsius has a pH of 7. Anything that has a pH less than 7 is called acidic, a pH above 7 is basic, or alkaline. Seawater has a pH just more than 8, whereas lemonade has a pH of approximately 3.

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PLEASE HELP!!! If you ever swam in a pool and your eyes began to sting and turn red, you felt the effects on an incorrect pH level. pH measures the concentration of hydronium ions and can be modeled by the function p(t) = −log10t. The variable t represents the amount of hydronium ions; p(t) gives the resulting pH level. Water at 25 degrees Celsius has a pH of 7. Anything that has a pH less than 7 is called acidic, a pH above 7 is basic, or alkaline. Seawater has a pH just more than 8, whereas lemonade has a pH of approximately 3.

Algebra
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1. Create a graph of the pH function either by hand or using technology. Locate on your graph where the pH value is 0 and where it is 1. You may need to zoom in on your graph. 2. The pool maintenance man forgot to bring his logarithmic charts, and he needs to raise the amount of hydronium ions, t, in the pool by 0.50. To do this, he can use the graph you created. Use your graph to find the pH level if the amount of hydronium ions is raised to 0.50. Then, convert the logarithmic function into an exponential function using y for the pH.
we have to use the definition of pH: \[\Large pH = - \log \left[ {{H^ + }} \right]\]

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more precisely: \[\Large pH = - {\log _{10}}\left[ {{H^ + }} \right]\]
you have to make a graph using a log scale, like this: |dw:1438104451173:dw|
namely you have to use a logarithmic scale for horizontal axis
for part B we can write this: \[\begin{gathered} y = - {\log _{10}}\left[ {{H^ + }} \right] = {\log _{10}}\left( {\frac{1}{{\left[ {{H^ + }} \right]}}} \right) \hfill \\ \hfill \\ \frac{1}{{\left[ {{H^ + }} \right]}} = {10^y} \Rightarrow \left[ {{H^ + }} \right] = {10^{ - y}} \hfill \\ \end{gathered} \]
\[\Large \begin{gathered} y = - {\log _{10}}\left[ {{H^ + }} \right] = {\log _{10}}\left( {\frac{1}{{\left[ {{H^ + }} \right]}}} \right) \hfill \\ \hfill \\ \frac{1}{{\left[ {{H^ + }} \right]}} = {10^y} \Rightarrow \left[ {{H^ + }} \right] = {10^{ - y}} \hfill \\ \end{gathered} \]
where \[\left[ {{H^ + }} \right]\] represents the amount of hydronium ions
would that be the x? and ph = y
yes!
you can use the logarithmic paper for part A

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