Determine the [H+] , [OH−] , and pH of a solution with a pOH of 9.19 at 25 °C.

Answer 1

Hello, Isaiah,

pOH = -log(OH^-]

pH = -log[H⁺] Note: [H⁺] is generally written as [H3O⁺]

That means if for a pOH of 9.19, we can find the [OH^-] concentration by these steps:

pOH = 9.19

[OH^-] = 10^-9.19

[OH^-] = 6.46x10^-10 M

For water at 25C, the equilibrium equation and constant is [H3O⁺]*[OH^-] =1x10^-14

[H3O⁺] =(1x10^-14)/*[OH^-]

[H3O⁺] =(1x10^-14)/*(6.46x10^-10)

[H3O⁺] = 1.55x10^-5 M

pH = -log(1.55x10^-5)

pH = 4.81

Using logs: Check the directions for your calculator for using the log functions. For logs to the base 10, the log is the value of the exponent, x. required to raise 10 (i.e., the base is 10) to the number specified. The log of 100, for example, is 2. That's because an exponent of 2 is required to raise 10 to the value specified (100). Going backwards, one can find the number being described by the log by raising 10 to the log value. If the log of a number is 3, you can calculate that number with 10³. Spreadsheets containing the log functionality are my favorite approach to working with logs. In Excel, one can also change the base of the log. If I see log₂ 2. I read it as reference to the base number (2) that is equivalent to 2² (4). 4 in the base 10 log system is 0.60206. So that 10^0.60206 = 4.

Bob