The pH of the monoprotic weak acid is 2.79.
<h3>What are weak acids?</h3>
The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.
The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c (monoprotic acid) = 0.33 M.
Ka = 1.2·10⁻⁶
[A⁻] = [H⁺] = x
[HA] = 0.33 M - x
Ka = [A⁻]·[H⁺] / [HA]
2. 6 × 10⁻⁶ = x² / (0.33 M - x)
Solve quadratic equation: [H⁺] = 0.000524 M.
pH = -log[H⁺]
pH = -log(0.000524 M)
pH = 2.79
Thus, the pH of the monoprotic weak acid is 2.79
To learn more about weak acids, refer to the below link:
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Answer : The rate constant at 785.0 K is, 
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= 
= rate constant at 
= ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature =
= final temperature =
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)
Therefore, the rate constant at 785.0 K is,
A chemical reaction is the process by which a CHANGE takes place (?)
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
