The pH of a 0.000001M solution of NaOH?

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Oct 21, 2017

Answer:

#sf(pH=7.04)#

Explanation:

This is a very dilute solution of NaOH so I would expect the pH to be 7 or very slightly above.

Because the concentration is so low we must also take into account the ions that are formed from the dissociation of water:

#sf(H_2OrightleftharpoonsH^++OH^-)#

#sf(K_w=[H^+][OH^-]=10^(-14)# at #sf(25^@C)#

If you do a thought experiment you can imagine adding a small amount of NaOH to water such that its concentration = #sf(10^-6color(white)(x)"mol/l")#.

The total concentration of #sf(OH^-)# ions is therefore #sf(10^(-6)+10^(-7)=1.1xx10^(-6)color(white)(x)"mol/l")#.

However this is not the concentration of #sf(OH^-)# ions at equilibrium so we cannot use this value to get the pH.

By adding those extra #sf(OH^-)# ions we have disturbed a system which is already at equilibrium. If you are familiar with Le Chatelier's Principle you will understand that such a system will react by opposing the change applied to it.

It will do this by shifting to the left thus reducing the concentration of #sf(OH^-)# ions to restore the value of #sf(K_w)#.

We need to set up an ICE table based on #sf("mol/l")#:

#sf(" "" "" "H_2O" "" "rightleftharpoons" "H^+" "" "+" "" "OH^-)#

#sf(I" "" "" "" "" "" "" "" "" "10^(-7)" "" "" "1.1xx10^(-6))#

#sf(C" "" "" "" "" "" "" "" "" "-x" "" "" "" "-x)#

#sf(E" "" "" "" "" "" "" "" "10^-7-x" "" "1.1xx10^(-6)-x)#

#:.##sf((1.1xx10^(-6)-x)(10^(-7)-x)=10^(-14))#

Multiplying out this gives:

#sf(x^2-(12xx10^(-7))x+10^(-14)=0)#

This is a quadratic equation which can be solved using the quadratic formula:

This gives:

#sf(x=(12xx10^(-7)+-sqrt(140xx10^(-11))))/(2)#

Ignoring the absurd root gives:

#sf(x=0.0839xx10^(-7))#

From the ICE table we get:

#sf([H^+]=1xx10^(-7)-0.0839xx10^(-7)=0.91607xx10^(-7)color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log(0.91607xx10^(-7))=7.04)#

As expected, very close to neutral.

This is why spillages of acid or alkali are irrigated with copious amounts of water to bring the pH towards neutrality.

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