document.write( "Question 540432: PH=-log[h]
\n" ); document.write( "if Ph=5.7
\n" ); document.write( "how we can calculate the log[h]
\n" ); document.write( " I need the answer with explanation.thank u
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Algebra.Com's Answer #353706 by KMST(5328) $\"\"$ $\"About$

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pH=-log[H^+] is calculated based on the molar concentration of the positively charged hydrogen ion. (The + or - charges for ions are written as superscripts, as if they were exponents. The p is always lowercase, never a capital letter, while H and O are always capital letters).
\n" ); document.write( "If pH=-log([H])=5.7 --> log([H])=-5.7 and [H]= $\"10%5E-5.7=2.0%2A10%5E-6=0.000002\"$
\n" ); document.write( "[H+]= $\"10%5E-5\"$ <--> pH=5, [H+]= $\"10%5E-6\"$ <--> pH=6, If the concentration is somewhere in between, so is the pH, but the higher [H+] the lower the pH.
\n" ); document.write( "pH=7 is neutral, pH<7 is acidic (excess H+ ions), and pH>7 is alkaline (also called basic).
\n" ); document.write( "BECAUSE I AM A CHEMIST AFTER ALL:
\n" ); document.write( "Water is dissociated to a very, very limited extent into H+ and OH- ions. There is an equilibrium between the whole molecules on one side and the ions on the other.
\n" ); document.write( "In pure water there are $\"10%5E-7\"$ moles per liter of each ion, and the product of those concentrations is $\"10%5E-14\"$ . The pH of that pure water would be
\n" ); document.write( " $\"pH=-log%2810%5E-7%29=7\"$ . That is called a neutral pH.
\n" ); document.write( "If you add to that water a compound that will provide more H+ or OH- ions, the concentrations will change, and some of the H+ and OH-ions will join to form more water molecules. The result will be new values for the molar concentrations of H+ and OH-, but those concentrations, multiplied will still be $\"10%5E-14\"$ . That number is a constant that is called the ionic product of water.
\n" ); document.write( "If you add an acid the concentration of H+ will increase. The pH and the concentration of OH- will decrease.
\n" ); document.write( "You may end up with [H+]=0.01= $\"10%5E-2\"$ (pH=2) and [OH-]= $\"10%5E-12\"$ .
\n" ); document.write( "If you add a base, like sodium hydroxide (NaOH), the concentration of OH- and the pH will increase, while the concentration of H+ will decrease.
\n" ); document.write( "You may end up with [OH-]=0.001= $\"10%5E-3\"$ , [H+]= $\"10%5E-11\"$ and pH=11 \n" ); document.write( "

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