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Now that you understand Strength and Concentration, it's time to learn how to calculate them!
The concentration is commonly stated in the form of molarity. If you have say, .1M of the acid HCl (Hydrochloric Acid), then that means you have .1 H+ ions. Similarly, if you have .1M of the base NaOH (Sodium Hydroxide), then you have .1 OH- ions.
5 equations are used to calculate the amount of H+ ions, OH- ions, pH level, and pOH level. To understand them, you'll need a graphing calculator (or use the one below).
Now that you understand Strength and Concentration, it's time to learn how to calculate them!
The concentration is commonly stated in the form of molarity. If you have say, .1M of the acid HCl (Hydrochloric Acid), then that means you have .1 H+ ions. Similarly, if you have .1M of the base NaOH (Sodium Hydroxide), then you have .1 OH- ions.
5 equations are used to calculate the amount of H+ ions, OH- ions, pH level, and pOH level. To understand them, you'll need a graphing calculator (or use the one below).
The 5 equations are as follows:
- pH + pOH = 14
- pH = -log ( [H+] )
- [H+] = 2ndlog, exponent: -pH
- pOH = -log ( [OH-] )
- [OH-] = 2ndlog, exponent: -pOH
"2ndlog" means that you press the "2nd" button on your personal graphing calculator, and then the "log" button. It then appears as the number 10, with an exponent waiting to be filled in. If you were looking to find the amount of H+ ions, you would plug in -pH as the exponent. If looking for the amount of OH- ions, you plug in -pOH as the exponent. HOWEVER: If you are using the calculator above, the button next to the "log" button (10^x) is the same thing as "2ndlog" on a TI-84 calculator (keep this in mind, for in the examples below, it is explained as if you are using a TI-84).
Lets do a few examples to understand the process. There are four ways you can solve a problem like this, depending on what numbers are provided for you (H+, OH-, pH, or pOH).
Example: pH is given
pH = 6.52
First, you should find the amount of H+ ions. To do this, you can use the 3rd equation above ( [H+] = 2ndlog, exponent: -pH ). Click the "2nd" button on your calculator, then "log", and then the "(-)" button (not the subtraction button!). Then fill in the pH as the exponent, which is 6.52, and press enter. You should get that
[H+] = 3.02E-7
Next, you should find the pOH level. To do this, use the first equation above (pH+pOH=14). Plug in your pH as 6.52, and you should get pOH=7.48
Lastly, you need to find the amount of OH- ions. Use the last equation above ( [OH-] = 2ndlog, exponent: -pOH ). You know from the last step that your pOH is 7.48, so you should click "2nd", then "log", then "(-)", plug in 7.48 as the exponent, and press enter. You should get
OH-=3.31E-8
Final Answers:
[H+] = 3.02E-7, [OH-] = 3.31E-8, pH = 6.52, pOH = 7.48
Example: pH is given
pH = 6.52
First, you should find the amount of H+ ions. To do this, you can use the 3rd equation above ( [H+] = 2ndlog, exponent: -pH ). Click the "2nd" button on your calculator, then "log", and then the "(-)" button (not the subtraction button!). Then fill in the pH as the exponent, which is 6.52, and press enter. You should get that
[H+] = 3.02E-7
Next, you should find the pOH level. To do this, use the first equation above (pH+pOH=14). Plug in your pH as 6.52, and you should get pOH=7.48
Lastly, you need to find the amount of OH- ions. Use the last equation above ( [OH-] = 2ndlog, exponent: -pOH ). You know from the last step that your pOH is 7.48, so you should click "2nd", then "log", then "(-)", plug in 7.48 as the exponent, and press enter. You should get
OH-=3.31E-8
Final Answers:
[H+] = 3.02E-7, [OH-] = 3.31E-8, pH = 6.52, pOH = 7.48
Not so bad right? If you think you get the concept or you just want to skip to the quiz already, click here. If you need more examples, keep reading.
Example: pOH is given
pOH = 10.78
First, you should find the amount of OH- ions. To do this, use the 4th equation above ( [OH-] = 2ndlog, exponent: -pOH). Like you have done before, press "2nd," "log," "(-)," and (instead of pH) plug in the pOH as the exponent (10.78). You should get
[OH-] = 1.66E-11
Next, you should find the pH. Use the first equation above (pH + pOH = 14). Plug in 10.78 as the pOH, and you should get pH = 3.22
Lastly, you need to find the number of H+ ions. To do this, use the third equation above ( [H+] = 2ndlog, exponent: -pH). Click "2nd," "log," "(-)," and plug in 3.22 (the pH). You should get [H+] = 6.026E-4
Final Answers:
[H+] = 6.026E-4, [OH-] = 1.66E-11, pH = 3.22, pOH = 10.78
Example: H+ is given
[H+] = 4.9E-4
First, you should find the pH level. To do this, use the second equation above (pH = -log ( [H+] ) ). On your calculator, click "(-)," then "log," and plug in 4.9E-4 for [H+] (for "E," click "2nd" then ",". Then close the parenthesis, and press enter. You should get pH = 3.31
Next, you should find the pOH level. Like you have before, use the first equation above (pH + pOH = 14), and plug in 3.31 as the pH. You should get pOH = 10.69
Lastly, you need to find the amount of OH- ions. To do this, use the last equation above ( [OH-] = 2ndlog, exponent: -pOH ). Click "2nd," "log," "(-)," and plug in 10.69 for pOH. You should get [OH-] = 2.04E-11
Final Answers: [H+] = 4.9E-4, [OH-] = 2.01E-11, pH = 3.31, pOH = 10.69
Example: OH- is given
[OH-] = 1.1E-4
First, you should find the pOH. To do this, use the 4th equation above (pOH = -log ( [OH-] ) ). Click "(-)," "log," plug in 1.1E-4 for [OH-], close the parenthesis, and press enter. You should get pOH = 3.96
Next, you should find pH level. To do this, use the first equation (pH + pOH = 14), and plug in 3.96 for pOH. You should get pH = 10.04
Lastly, you need to find the number of H+ ions. To do this, you use the third equation above ( [H+] = 2ndlog, exponent: -pH). Click "2nd," "log," "(-)," and plug in 10.04 for the pH. You should get [H+] = 9.12E-11
Final Answers: [H+] = 9.12E-11, [OH-] = 1.1E-4, pH = 10.04, pOH = 3.96
The hardest part is over! Now onto a lab!
pOH = 10.78
First, you should find the amount of OH- ions. To do this, use the 4th equation above ( [OH-] = 2ndlog, exponent: -pOH). Like you have done before, press "2nd," "log," "(-)," and (instead of pH) plug in the pOH as the exponent (10.78). You should get
[OH-] = 1.66E-11
Next, you should find the pH. Use the first equation above (pH + pOH = 14). Plug in 10.78 as the pOH, and you should get pH = 3.22
Lastly, you need to find the number of H+ ions. To do this, use the third equation above ( [H+] = 2ndlog, exponent: -pH). Click "2nd," "log," "(-)," and plug in 3.22 (the pH). You should get [H+] = 6.026E-4
Final Answers:
[H+] = 6.026E-4, [OH-] = 1.66E-11, pH = 3.22, pOH = 10.78
Example: H+ is given
[H+] = 4.9E-4
First, you should find the pH level. To do this, use the second equation above (pH = -log ( [H+] ) ). On your calculator, click "(-)," then "log," and plug in 4.9E-4 for [H+] (for "E," click "2nd" then ",". Then close the parenthesis, and press enter. You should get pH = 3.31
Next, you should find the pOH level. Like you have before, use the first equation above (pH + pOH = 14), and plug in 3.31 as the pH. You should get pOH = 10.69
Lastly, you need to find the amount of OH- ions. To do this, use the last equation above ( [OH-] = 2ndlog, exponent: -pOH ). Click "2nd," "log," "(-)," and plug in 10.69 for pOH. You should get [OH-] = 2.04E-11
Final Answers: [H+] = 4.9E-4, [OH-] = 2.01E-11, pH = 3.31, pOH = 10.69
Example: OH- is given
[OH-] = 1.1E-4
First, you should find the pOH. To do this, use the 4th equation above (pOH = -log ( [OH-] ) ). Click "(-)," "log," plug in 1.1E-4 for [OH-], close the parenthesis, and press enter. You should get pOH = 3.96
Next, you should find pH level. To do this, use the first equation (pH + pOH = 14), and plug in 3.96 for pOH. You should get pH = 10.04
Lastly, you need to find the number of H+ ions. To do this, you use the third equation above ( [H+] = 2ndlog, exponent: -pH). Click "2nd," "log," "(-)," and plug in 10.04 for the pH. You should get [H+] = 9.12E-11
Final Answers: [H+] = 9.12E-11, [OH-] = 1.1E-4, pH = 10.04, pOH = 3.96
The hardest part is over! Now onto a lab!