The answer’s in the picture
Answer:
Increase in temperature increases the average kinetic energy of the reactant molecules, hence lowering the activation energy of the reaction.
Answer:
28.0mL of the 0.0500M NaOH solution
Explanation:
<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>
<em />
The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:
H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
<em />
You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.
the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:
0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>
To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:
1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =
<h3>28.0mL of the 0.0500M NaOH solution</h3>
Answer:
pH = 6.82
Explanation:
To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:
- pH = pKa + log
![\frac{[NaOCl]}{[HOCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl%5D%7D%7B%5BHOCl%5D%7D)
We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:
- pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
- [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
- pH = 7.54 + log
By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:
- pH = 7.54 + log
![\frac{[NaOCl-HCl]}{[HOCl+HCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl-HCl%5D%7D%7B%5BHOCl%2BHCl%5D%7D)
- pH = 7.54 + log
