We should apply Boyle's Law here given initial pressure, initial volume and final volume.
P1V1= P2V2
(6.5 atm) (13 L) = P2 (3.3 L)
Solve for P2 on your calculator and that should get you to the answer.
Answer:
0.3229 M HBr(aq)
0.08436M H₂SO₄(aq)
Explanation:
<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>
<em />
Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).
NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)
When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>
<em />
Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).
2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)
When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:
<span>H2C=O---------H-OH </span>
<span>But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. </span>
<span>Formaldehyde in aqueous solution is in equilibrium with its hydrate. </span>
<span>H2C=O + H2O <-----------------> H2C(OH)2</span>
The formula we will use is:
pH = 0.5 (pKa – log C)
where C is the concentration of sulfathiazole in molarity
or mol / L
Since all values are given, we can compute directly for
pH:
pH = 0.5 (7.12 – log 0.002)
<span>pH = 4.91</span>
Answer:
Explanation:
<u>1) Data:</u>
a) Hypochlorous acid = HClO
b) [HClO} = 0.015
c) pH = 4.64
d) pKa = ?
<u>2) Strategy:</u>
With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.
<u>3) Solution:</u>
a) pH
b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)
c) Equilibrium constant: Ka = [ClO⁻] [H₃O⁺] / [HClO]
d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M
e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M
f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46