anonymous
  • anonymous
PLEASE HELP!!! 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 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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.
anonymous
  • anonymous
Im not quite sure how to set up the equation; like where the "0" and the "1" fit into the pH equation... someone please help
anonymous
  • anonymous
we have to use the definition of pH: \[\Large pH = - \log \left[ {{H^ + }} \right]\] 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 that would be the x? and ph = y You can use the logarithmic paper for part A @Luck8176 k

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