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A titration is performed making use of 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 that 0.100 M NaOH.

Answer:

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

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

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

You are watching: Hcl(aq)+naoh(aq)→nacl(aq)+h2o(l)


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

The mole of NaOH added from the buret are;

*

The mole of HCl in the flask initially 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 go come completion. This method that the reactant in the smallest amount will totally react. At this point in the titration there are fewer moles of base contrasted to acid. So all the base reacts. Including this to 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 the each types can be derived just together it always has been in the other ICE tables, by adding the adjust to the early stage 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 last row there is mountain remaining and also all the base has reacted. Salt has been formed.

To calculation the pH we must recognize the kind of device we have. To execute that us look at the final row and also see what types are present. In this case there are moles that a solid acid (HCl) and its salt (NaCl). But the salt the a solid acid and also a strong base walk not impact the pH of water, so we left with just moles of HCl in the mixture.

To calculate the pH the the equipment we must determine the concentration that HCl in the mixture. The concentration is acquired by splitting the volume that the solution right into the moles of HCl. The volume of the systems is 50.0 mLs (40.0 mLs of HCl and also 10.0 mLs that NaOH).

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So the is;

*


Solving because that the pH that a strong acid;

pH = - log

Since strong acids totally dissociate the = = 0.0600 M

pH = - log 0.0333

pH = 1.48

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

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