Question #8eff7

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#