pH is killing me

[nooshnoosh95]

2 questions
1. 10mL of 0.100 mol L-1 NaOH solution is diluted by addition of 990mL of distilled water. The resulting pH is?

and

2. How much water must be added to 1L of a solution of a strong base of pH 13 in order to decrease the pH to 12?

please show me how to work these out
thanks in advance

[charmanderp]

Initial pH is 14-(-log(0.1))=13. It's being diluted by a factor of 10^2 so the new pH is 11.

Alternatively C1=0.1 and C2=(C1*V1)/V2=0.001M

14-(-log(0.001)=11.

As above, a change in pH by one down (when greater than 7) or up (when less than 7) is a dilution by a factor of 10. So 1*10-1=9L

[nooshnoosh95]

thanks babe

[Tonychet2]

i had trouble with ph and i still cant do it quick but this method always works (for me anyway)

u know that [H+] * [OH-] = 10^-14 and ph = -log[H+]

there are different situations but ill list a few

to solve for pH
if given concentration of [H+] u just sub it into the formula to get pH
if given concentration of [OH-] u use basic algebra to rearrange for [H+] so it would be [H+] = 10^-14 divide by [OH-] , then you can sub into ph = -log[H+] to get your answer

to solve for [H+]
when given pH u rearrange to get [H+] = 10^-pH

to solve for [OH-]
when given pH u rearrange to get [H+] from 10^-pH

now you have the concentration of [H+] u can solve for [OH-] by letting [H+] * [OH-] = 10^-14, therefore [OH-] = 10^-14 divided by [H+]