pH and pOH Calculations
Now we know that pH directly corresponds with the molarity of H+ ions as explained on the previous page. The same can be said with the relationship between pOH and OH- ions. pOH is another scale that measures the pH of a solution, except the pOH scale is opposite the pH scale. As a result 0-6 is basic, 7 is still neutral and 8-14 is acidic. The pOH scale is dependent on the molarity of OH- ions. pOH is dependent on the [OH-] as the pH value is dependent on the [H+] value. But what is the exact relationship between all of these values one might ask? There are formulas that link any of the four values (pH, pOH, [H+], [OH-]) so that when a scientist has only one of the values of a solution, he or she will be easily able to find the other three. The pOH value is dependent on the [OH-] and vice versa. As a result all of the equations relate [OH-] and pOH and pH and [H+] are separate. There is the same relationship between pH and [H+], pH is dependent on the [H+] and vice versa. As a result all of the equations relate [H+] and pH and pOH and [OH-] are separate. Then the pOH is achieved by subtracting the pH from 14 and it is also works the in the opposite way too because the pH is found by subtracting the pOH from 14. Here are the general equations (their uses will be explained below):
Lets say one knows the pH that a solution, which is 2.7. From here, one can put 2.7 into the equation to find the [H+]:
There will be circumstances where one will only have molarity of a specific solution. To proceed, one will have to figure out if the solution is an acid or base. If the solution is an acid, then one would know that the molarity is referring to the [H+]. If the solution is a base, then one would know that the molarity is referring to the [OH-]. One would distinguish the acid where the solution is an acid or a base by using the Arrhenius Theory and looking if the chemical possesses either an H+ ion or an OH- ion. For example, if one knows that they possess 0.0407 M of NaOH, then they would know that this is the [OH-] and the solution would proceed from there.
Lets say one has 0.0002 M of HCl find [H+], [OH-], pH and pOH:
- pH= -log([H+])
- pOH= -log([OH-])
- pH= 14- pOH
- pOH= 14- pH
- [H+]= 2nd log(-pH) ( in order to fulfill this on a graphing or scientific calculator: One clicks the second button on the calculator then the log button, and then the little negative that is to the right of the decimal point and then types the pH)
- [OH-]= 2nd log(-pOH) ( in order to fulfill this on a graphing or scientific calculator: One clicks the second button on the calculator then the log button, and then the little negative that is to the right of the decimal point and then types the pOH)
Lets say one knows the pH that a solution, which is 2.7. From here, one can put 2.7 into the equation to find the [H+]:
- [H+]= 2nd log(-pH) so our equation is [H+]= 2nd log(-2.7). This equals 1.995e-3 which is the value of [H+}.
- From there, one find the pOH by subtracting the pH from 14--> 14-2.7= 11.3. The equation equals 11.3 so the pOH= 11.3
- Now one can find the [OH-] by putting the pOH into the same equation that was used to find the pH: [OH-]= 2nd log(-pOH) --> [OH-]= 2nd log(-11.3) . The equation equals 5.012 x 1e-12 So this means that the [OH-]= 5.012e-12
- pOH= -log([OH-]) so our equation is pOH= -log(5.24e-13). This equals 12.28 which the pOH.
- From there, one can find the pH by subtracting the pOH by 14, so our equation is 14-12.28. This equals 1.72, so the pH is 1.72.
- Now one can find the [H+] by putting the pH into the equation: pH= 2nd log (-pH). So our equation is 2nd log(-1.72). This equals 1.905e-2 so [H+]= 1.905e-2.
- pH= -log([H+]) so our equation is pH= -log(3.39e-3). This equation equals 2.4698 which means the pH= 2.4698
- From there, one can find the pOH by subtracting the pH from 14. So our equation is 14-2.4698 and it equals 11.5302. This means that the pOH= 11.5302
- Next one can find the [OH-} by putting the pOH into the equation: [OH-]= 2nd log(-pOH) so our equation is [OH-]= 2nd log( -11.5302). This equation equals 2.9499e-12 so the [OH-]= 2.9499e-12
- [OH-]= 2nd log(-pOH) so our equation is [OH-]= 2nd log(-5.21). This equation equals 6.166e-6 so the [OH-]= 6.166e-6
- From there, one can find the pH by subtracting the pOH by 14. So our equation is pH= 14-5.21. This equation equals 8.79. This means that the pH= 8.79
- Next one can find the [H+] by putting the pH into the equation: [H+]= 2nd log(-pH). So our equation is [H+]= 2nd log(-8.79). This equation equals 1.622e-9 so the [H+]= 1.622e-9
There will be circumstances where one will only have molarity of a specific solution. To proceed, one will have to figure out if the solution is an acid or base. If the solution is an acid, then one would know that the molarity is referring to the [H+]. If the solution is a base, then one would know that the molarity is referring to the [OH-]. One would distinguish the acid where the solution is an acid or a base by using the Arrhenius Theory and looking if the chemical possesses either an H+ ion or an OH- ion. For example, if one knows that they possess 0.0407 M of NaOH, then they would know that this is the [OH-] and the solution would proceed from there.
Lets say one has 0.0002 M of HCl find [H+], [OH-], pH and pOH:
- One can see that the molarity above is [H+] because the chemical that has the molarity above is an acid. We are able distinguish the chemical as an acid because the chemical contains an H+ ion, which as the Arrhenius Theory states, all acids dissociate in water to create H+ ions, meaning that all acids must contain at least one H+ ion. So from here one knows that [H+]= 2e-4
- From there one can find the pH by putting the [H+] into the equation: pH= -log([H+). So our equation is pH= -log (2e-4). This equation equals 3.698, which means that pH= 3.698.
- From there one can find the pOH by subtracting the pH from 14, so one will have the equation: 14-3.69. This equation equals 10.301, which means pOH= 10.301.
- Next one can find the [OH-] by putting the pOH into the equation: 2nd log (-pOH). So our equation is [OH-]= 2nd log (-10.301). This equation equals 4.898e-11, which means that [OH-]= 4.898e-11