Question

In part 1 of lab 2 you will make and standardize a solution of naoh(aq). suppose in the lab you measure the solid naoh and dissolve it into 100.0 ml of water. you then measure 0.2005 g of khp (kc8h5o4, 204.22 g/mol) and place it in a clean, dry 100-ml beaker, and then dissolve the khp in about 25 ml of water and add a couple of drops of phenolphthalein indicator. you titrate this with your naoh(aq) solution and find that the titration requires 9.82 ml of naoh(aq).

Answer 1

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Answer 2

Answer:

0.100 M

Explanation:

KHP is used to standardize the NaOH solution, according to the following neutralization reaction.

KC₈H₅O₄ + NaOH → NaKC₈H₄O₄ + H₂O

We can establish the following relations.

• The molar mass of KHP is 204.22 g/mol.
• The molar ratio of KHP to NaOH is 1:1.

The moles of NaOH that reacted with 0.2005 g of KHP are:

$0.2005gKHP.\frac{1molKHP}{204.22gKHP} .\frac{1molNaOH}{1molKHP} =9.818 \times 10^{-4} molNaOH$

The molarity of NaOH is:

$M=\frac{9.818 \times 10^{-4} mol}{9.82 \times 10^{-3}L} =0.100M$