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
The solubility is the guide to the maximum amount of solute that can be dissolved in a certain amount of solvent at a certain temperature to make a saturated solution. Any amount less than this would result to unsaturated, while any amount more would result to saturated.
6 g/(50 mL * 1 L/1000 mL) = 120 g/L
Since it is less than the solubility of 125 g/L, then <em>this solution is unsaturated</em>.
Answer:
Hi !
Here is your answer !
Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.
Explanation:
Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place.
Thank You !
Answer:
Explanation:
MM of I2 = 2 (127 g) = 254 g/mol
0.065 mol I2 x 254g I₂/ 1 mol I₂ = 16.5 g I2