A solution's ph is a measure of what? strength or weakness acid dissociation constant hydronium ion concentration molarity perce
1 answer:
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The pH value of the solution is mathematically given as
pH=2.35
<h3>What pH value of the solution?</h3>Question Parameters:
pH during the titration of 20.00 mL of 0.1000 M dimethylamine,
with 0.1000 M HCl(aq) after 21.23 mL of the acid
Generally, the equation for the Chemical Reaction is mathematically given as
(CH3)2NH(aq), +Hcl ---> <---- (CH3)2NH2Cl(aq)
Therefore
HCL=0.00444M
WHere
HClaq--->H+(aq)+Cl-(aq)
Hence
H+=0.00444M
pH= -log{H+}
pH=log(0.00444)
pH=2.35
For more information on Chemical Reaction
Answer: There are 6.9 mol of are required to react completely with 2.30 mol of S.
Explanation:
The given reaction equation is as follows.
Here, 1 mole of S is reaction with 3 moles of which means 1 mole of S requires 3 moles of
.
Therefore, moles of required to react completely with 2.30 moles S are calculated as follows.
Thus, we can conclude that there are 6.9 mol of are required to react completely with 2.30 mol of S.
Answer:
pH = 8.0
Explanation:
First, we have to calculate the moles of NaOH.
Let's consider the balanced equation.
C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O
The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.
The concentration of C₂H₃O₃Na is:
C₂H₃O₃Na dissociates according to the following equation:
C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)
C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.
C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻
If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:
pKa + pKb = 14
pKb = 14 -3.9 = 10.1
10.1 = -log Kb
Kb = 7.9 × 10⁻¹¹
We can calculate [OH⁻] using the following expression:
[OH⁻] = √(Kb.Cb) <em>where Cb is the initial concentration of the base</em>
[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M
Now, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0
pH + pOH = 14
pH = 14 - pOH = 14 - 6.0 = 8.0