Answer:
<h2>4.0 </h2>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ { H}^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7B%20H%7D%5E%7B%2B%7D%5D)
From the question we have
We have the final answer as
<h3>4.0</h3>
Hope this helps you
Answer:
P₂ = 13.9 atm (3 sig. figs.)
Explanation:
The pressure (P), Volume (V) relationship with Temperature (T) & mass (n) held constant is an inverse proportionality. That is Boyles Law ...
P ∝ 1/V => P = k/V => k = P·V
For two pressure-volume conditions, the proportionality constant (k) remains constant where k₁ = k₂ and P₁·V₁ = P₂·V₂ => P₂ = P₁·V₁/V₂
Given:
P₁ = 1.31 atm.
V₁ = 5.51 L
P₂ = ?
V₂ = 0.520 L
V₂ = (1.31 atm)(5.51L)/(0.520L) = 13.88096154 atm (calc. ans.) = 13.9 atm (3 sig. figs.)
The volume of the buffer solution having a ph value is calculated by henderson's hasselbalch equation.
Buffer solution is water based solution which consists of a mixture containing a weak acid and a conjugate base of the weak acid. or a weak base and conjugate acid of a weak base.it is a mixture of weak acid and a base. The pH of the buffer solution is determined by the expression of the henderson hasselbalch equation.
pH=pKa + log [salt]/[acid]
Where, pKa =dissociation constant , A- = concentration of the conjugate base, [HA]= concentration of the acid. Here, a buffer solution contains 0.403m acetic acid and 250 ml is added in order to prepare a buffer with a ph of 4.750. Putting all the values in the henderson hasselbalch equation we find the pH of the buffer solution.
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The answer is D. I did that and i got it right.
pH of 0.40M triethylammonium chloride is 5.90.
<h3>What is pH?</h3>
A solution's acidity may be determined by looking at its pH, which is a measurement of hydrogen ion concentration. Pure water slightly separates into ions with roughly equal amounts of hydrogen and hydroxyl (OH) ions. [H+] is 107 for a neutral solution, or pH = 7.
<h3>Given : </h3>
Concentration of triethylammonium chloride = 0.40M
pH = ?
<h3>Solution: </h3>
(CH3CH2)3NHCl ------> (CH3CH2)3NH⁺ + Cl⁻
(CH3CH2)3NH⁺ will react with water to give H3O⁺ .
(CH3CH2)3N will have a Kb = 5.2 x 10 ^(-4)
Kw = Kb x Ka
=> Ka = Kw / Kb = 10^(-14) / 5.2 x 10 ^(-4)
=> Ka = 1.92 x 10^(-11)
so by the reaction we have ,
Ka = x²/(0.40 - x)
=> x = 1.2393 x 10 ^(-6)
now, pH = -log( [H3O⁺]) = - log ( 1.2393 x 10 ^(-6)) = 5.906
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