Answer:
![pH=4.56](https://tex.z-dn.net/?f=pH%3D4.56)
Explanation:
Hello there!
In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:
![n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol](https://tex.z-dn.net/?f=n_%7Bacid%7D%3D%5Cfrac%7B0.2g%7D%7B234g%2Fmol%7D%3D0.000855mol%5C%5C%5C%5Cn_%7Bbase%7D%3D%20%5Cfrac%7B0.2g%7D%7B256g%2Fmol%7D%3D0.000781mol)
And the concentrations are:
![[acid]=0.000855mol/0.025L=0.0342M](https://tex.z-dn.net/?f=%5Bacid%5D%3D0.000855mol%2F0.025L%3D0.0342M)
![[base]=0.000781mol/0.025L=0.0312M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.000781mol%2F0.025L%3D0.0312M)
Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:
![pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56](https://tex.z-dn.net/?f=pH%3D-log%282.5x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.0312M%7D%7B0.0342M%7D%20%29%5C%5C%5C%5CpH%3D4.60-0.04%5C%5C%5C%5CpH%3D4.56)
Best regards!
<span>The structural formula of 2-methylbutan-2-ol is in Word document below.
</span>2-methyl-2-butanol is organic compound and belongs to alcohols. Hydroxyl <span>functional group is on second saturated carbon atom of butane and also methyl group (-CH</span>₃) is on second saturated carbon atom of main chain (butane).<span>
</span>
We know that there are 100 cm in 1 m, so we can use this to convert to meters:
![(\frac{2.41x10^{2}cm}{1})*(\frac{1m}{100cm} )=2.41m](https://tex.z-dn.net/?f=%20%28%5Cfrac%7B2.41x10%5E%7B2%7Dcm%7D%7B1%7D%29%2A%28%5Cfrac%7B1m%7D%7B100cm%7D%20%29%3D2.41m%20)
Therefore we know that
cm is equal to 2.41 m.