Determine the [ H3O+] and [OH-] of 1.0 M NaCl if the pH is 6.04

Question

anonymous · Answer

Since your pH=6.04
Therefore, you can easily solve for the [H3O+] or [H+] (since both are exactly the same thing) using the antilog, or raising 10 to the negative power of the pH.
$$[H3O+]=10^{-pH}=10^{-6.04}=9.12011\times10^{-7}M$$ 
Next, to solve for the OH- concentration, remember that:
$$[OH-]=Kw \div[H3O+]$$ -------OR------- $$[OH-]=10^{-(14-pH)}$$
$$remember Kw=1\times10^{-14}$$
Kw is the ionization constant of water, which basically shows that(ALWAYS) $$[H3O+]\times[OH-]=Kw=1\times10^{-14}$$
So,
$$[OH-]=1\times10^{-14}\div(9.12011\times10^{-11}M)=1.09648\times10^{-8}M$$
$$[OH-]=10^{-(14-6.04)}=1.90648\times10^{-8}M$$
I solved for [OH-] two different ways, but still got the same answer, which basically shows I was doing the same equation but with different steps.
Hope this helps!