Answer:
The pH of 0.1 M BH⁺ClO₄⁻ solution is <u>5.44</u>
Explanation:
Given: The base dissociation constant: 
= 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M
Also, water dissociation constant: 
= 1 × 10⁻¹⁴
<em><u>The acid dissociation constant </u></em>(
)<em><u> for the weak acid (BH⁺) can be calculated by the equation:</u></em>
<em><u>Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:</u></em>
Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+
Initial: 0.1 M x x
Change: -x +x +x
Equilibrium: 0.1 - x x x
<u>The acid dissociation constant: </u>![K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5Cleft%20%5BB%20%5Cright%20%5D%20%5Cleft%20%5BH_%7B3%7DO%5E%7B%2B%7D%5Cright%20%5D%7D%7B%5Cleft%20%5BBH%5E%7B%2B%7D%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B%28x%29%28x%29%7D%7B%280.1%20-%20x%29%7D%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%7D%7B0.1%20-%20x%7D)
<u>Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M</u>
Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44
<u>Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44</u>
Answer:
с
Explanation:
the first quantum number of an electron gives the information about the energy level the electron is in
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer:
806.3g
Explanation:
Given parameters:
Number of moles of silver nitrate = 4.85mol
Unknown:
Mass of silver chromate = ?
Solution:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃
To solve this problem, we work from the known to the unknown;
- The known specie here is AgNO₃ ;
From the balanced chemical equation;
2 moles of AgNO₃ will produce 1 mole of Ag₂CrO₄
4.85 moles of AgNO₃ will produce 
= 2.43moles of Ag₂CrO₄
- Mass of silver chromate produced;
mass = number of moles x molar mass
Molar mass of Ag₂CrO₄
Atomic mass of Ag = 107.9g/mol
Cr = 52g/mol
O = 16g/mol
Input the parameters and solve;
Molar mass = 2(107.9) + 52 + 4(16) = 331.8g/mol
So,
Mass of Ag₂CrO₄ = 2.43 x 331.8 = 806.3g
