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A titration is performed utilizing by including 0.100 M NaOH to 40.0 mL of 0.1 M HCl.

b") calculation the pH after enhancement of 20.0 mL the 0.100 M NaOH.

Answer:

When a solid base prefer NaOH is added to a solid acid prefer HCl a neutralization reaction occurs,

NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(aq)

The K for this reaction is very large, therefore the reaction goes come completion. Come determine how much reaction wake up we require to collection up an ice cream "like" table.

You are watching: Naoh+hcl=nacl+h2o


Determine the mole of NaOH (that to be added) and the mole of HCl (present initially);

The mole of NaOH included from the buret are;

*

The moles of HCl in the flask originally are;

*

So the "ICE table" currently looks like;


NaOH(aq)

+ HCl(aq)

--->

NaCl(aq)

+ H2O(aq)

I

0.00200 mol

0.00400 mol

0

-

C

F

Since K because that the reaction is large, the reaction will certainly go come completion. This method that the reactant in the smallest amount will fully react. In ~ this allude in the titration there room fewer mole of base compared to acid. So all the base reacts. Including this come the "ICE table" us have;


NaOH(aq)

+ HCl(aq)

--->

NaCl(aq)

+ H2O(aq)

I

0.00200 mol

0.00400 mol

0

-

C

-0.00200 mol

-0.00200 mol

+0.00200 mol

-

F


THe final amount of each species can be derived just together it constantly has been in the various other ICE tables, by adding the change to the early amount.


NaOH(aq)

+ HCl(aq)

--->

NaCl(aq)

+ H2O(aq)

I

0.00200 mol

0.00400 mol

0

-

C

-0.00200 mol

-0.00200 mol

+0.00200 mol

-

F

0

0.00200 mol

+0.00200 mol


Focusing on the final row over there is mountain remaining and all the base has actually reacted. Salt has been formed.

To calculate the pH we should recognize the form of system we have. To carry out that we look in ~ the last row and also see what varieties are present. In this case there room moles the a strong acid (HCl) and also its salt (NaCl). Yet the salt the a solid acid and also a solid base does not effect the pH that water, so we left with only moles that HCl in the mixture.

To calculation the pH that the solution we must recognize the concentration that HCl in the mixture. The concentration is obtained by splitting the volume of the solution into the moles of HCl. The volume the the equipment is 50.0 mLs (40.0 mLs the HCl and also 10.0 mLs that NaOH).

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So the is;

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Solving because that the pH the a strong acid;

pH = - log in

Since strong acids totally dissociate the = = 0.0600 M

pH = - log 0.0333

pH = 1.48

We deserve to see this suggest on the titration curve (as suggest 1b").

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