The question is incomplete, here is the complete question:
Calculate the pH of a solution prepared by dissolving 0.370 mol of formic acid (HCO₂H) and 0.230 mol of sodium formate (NaCO₂H) in water sufficient to yield 1.00 L of solution. The Ka of formic acid is 1.77 × 10⁻⁴
a) 2.099
b) 10.463
c) 3.546
d) 2.307
e) 3.952
<u>Answer:</u> The pH of the solution is 3.546
<u>Explanation:</u>
We are given:
Moles of formic acid = 0.370 moles
Moles of sodium formate = 0.230 moles
Volume of solution = 1 L
To calculate the molarity of solution, we use the equation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[HCOONa]}{[HCOOH]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5BHCOONa%5D%7D%7B%5BHCOOH%5D%7D%29)
= negative logarithm of acid dissociation constant of formic acid = 3.75
![[HCOOH]=\frac{0.370}{1}](https://tex.z-dn.net/?f=%5BHCOOH%5D%3D%5Cfrac%7B0.370%7D%7B1%7D)
pH = ?
Putting values in above equation, we get:

Hence, the pH of the solution is 3.546
(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6
I think the word might be a “troll”
For a given peak intensity of radiation of a star that occurs at a wavelength of 2 nanometers, this is located at the spectral band of an X-ray. An X-ray's wavelength typically goes from 0.1 nano meters to 10 nano meters. Given that, the wavelength of the star fits perfectly into the range of an X-ray