Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
Answer:
B. Gas formation
Explanation:
When you combine baking soda and vinegar - which I'm assuming happened to create this reaction - it forms carbon dioxide, a gas.
A water molecule, because of its shape, is a polar molecule. That is, it has one side that is positively charged and one side that is negatively charged. The molecule is made up of two hydrogen atoms and one oxygen atom. The bonds between the atoms are called covalent bonds, because the atoms share electrons.
<u>Given:</u>
Volume of Na2CO3 = 250 ml = 0.250 L
Molarity of Na2CO3 = 6.0 M
Volume of CaF2 = 750 ml = 0.750 L
Molarity of CaF2 = 1.0 M
<u>To determine:</u>
The mass of CaCO3 produced
<u>Explanation:</u>
Na2CO3 + CaF2 → CaCO3 + 2NaF
Based on the reaction stoichiometry:
1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3
Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles
Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles
CaF2 is the limiting reagent
Thus, # moles of CaCO3 produced = 0.750 moles
Molar mass of CaCO3 = 100 g/mol
Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g
Ans: Mass of CaCO3 produced = 75 g
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
