mass of proton mp=1.672621898e-27 kg and 1 amu=1.66053904e-27kg so it’s just divition (1.672621898e-27 / 1.66053904e-27) =1.007276 amu
Answer:
About 0.0940 M.
Explanation:
Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

Therefore, the molarity of the original solution was:
![\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%5B%20%5Ctext%7BHA%7D%5Cright%5D%20%3D%20%5Cfrac%7B0.00188%5Ctext%7B%20mol%7D%7D%7B20.00%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%3D%200.0940%5Ctext%7B%20M%7D)
In conclusion, the molarity of the unknown acid is about 0.0940 M.
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