Question #8eff7
1 Answer
Explanation:
Well, all you really have to do here is use the equation given to you. The pH of a solution can be calculated by using
#color(purple)(bar(ul(|color(white)(a/a)color(black)("pH" = - log(["H"^(+)]))color(white)(a/a)|)))#
Before plugging in the value given to you for the concentration of hydrogen ions, it's worth mentioning that because you're dealing with the negative log, a higher concentration of hydrogen ions will result in a lower pH.
Likewise, a lower concentration of hydrogen ions will result in a higher pH.
In your case, you know that
#["H"^(+)] = 6.3 * 10^(-3)"M"#
The pH of the solution will be
#"pH" = - log(6.3 * 10^(-3))#
#"pH" = 2.20#
Here's a cool thing to remember about logs and pH. You can manipulate the above equation to write
#"pH" = - [log(6.3) + log(10^(-3))]#
#"pH" = - [log(6.3) + (-3) * log(10)]#
#"pH" = -[log(6.3) - 3]#
#"pH" = 3 - log(6.3)#
#"pH" = 2.20#
Notice that for
#["H"^(+)] = 6.3 * 10^(-color(blue)(3))#
you have
#"pH" = color(blue)(3) - log(6.3)#
This allows you to estimate the pH of a solution just by looking at the concentration of hydrogen ions. For example, if
#["H"^(+)] = 1.5 * 10^(-color(red)(4))#
you can say that you have
#"pH" = color(red)(4) - log(1.5) < 4#
