Answer:
[Na₂CO₃] = 0.094M
Explanation:
Based on the reaction:
HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)
It is possible to find pH using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base, [CO₃²⁻] = [Na₂CO₃] and [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.
pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>
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Replacing these values:
10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]
<em> [Na₂CO₃] = 0.094M</em>
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Answer:
Fluorine
Explanation:
Fluorine has the greatest attraction for electrons in any bond that it forms. The attraction of an atom for shared electrons is called its electronegativity.
WAit I just have one tiny little question did your mom dad grandma grandpa helpful I with the question first
In order to synthesize a complex organic molecule, the
chemist should at least illustrate or imagine the bonds that are in need to be
cut down or to separate in order to obtain the compound that can be easily
changed.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
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