Using your data from the experiment, calculate the initial moles of HCl that you started with. ( for this problem you will use your initial concentration of HCl and your 50 mL volume to solve).
| Trial 1 | trail 2 | |
| Mass of KHP (KC8H5O4) (g) | .29 g | .21 g |
| Moles of KHP | .000422 | .00422 |
| Moles of NaOH | .01M | .01M |
| Volume of NaOH | .0247 | .0222 |
| Molarity of NaOH | .0575 | .0464 |
| Average molarity of NaOH | .0516 M | |
Number of mole s= molarity * volume
Average volume of NaOH = 0.0247 +0.022
= 0.02345 mlNaOH
mole of NaOH = molarity * volume in L
= 0.1* 0.02345/1000
= 2.345*10^-6 mole NaOH
The recation between HCl and NaOH is as follows:
HCl + NaOH= NaCl+ H2O
Therefore 2.345*10^-6 mole NaOH reacts with 2.345*10^-6 mole of HCl
Molarity = number of moles / volume in L
Here volume of HCl= 50 ml
0.050 L
Thus
Molarity = 2.345*10^-6 mole HCl/0.050 L
= 4.69*10^-5 M
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