The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL
<h3>Balanced equation </h3>
CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O
From the balanced equation above,
- The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the volume of NaOH</h3>
- Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
- Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
- Molarity of base, NaOH (Mb) = 0.242 M
- Volume of base, KOH (Vb) =?
MaVa / MbVb = nA / nB
(0.204 × 46.79) / (0.242 × Vb) = 1
Cross multiply
0.242 × Vb = 0.204 × 46.79
Divide both side by 0.242
Vb = (0.204 × 46.79) / 0.242
Vb = 39.44 mL
Thus, the volume of NaOH needed for the reaction is 39.44 mL
Learn more about titration:
brainly.com/question/14356286
Answer: 8.3 J
Explanation:
We have the following measurement:
Rearranging the units:
Since 1 Newton is 
:
Since 1 Joule is 
:
This is the simplest form possible
Answer:
CH3
|
CH3- C H -CH2-CH2- CH - CH2-CH2-CH3
|
CH
/ \
CH3 CH3
Explanation:
Octan
C-C-C-C-C-C-C-C
Metyl
CH3 -
Isopropyl
CH3
/
- CH
\
CH3
2-metil-5-isopropiloctan
CH3
|
CH3- C H -CH2-CH2- CH - CH2-CH2-CH3
|
CH
/ \
CH3 CH3
The solution changed color because the substances are not neutral.
<h3>
pH</h3>
Chemical substances have different concentrations of the hydrogen cation, called PH.
The higher the pH, the more basic the substance, and the lower the more acidic.
Bromothymol blue is a pH indicator that changes its color according to the pH of the substance, yellow for acid, blue for basic and green for neutral.
In the case of the reactions in question, we have the release of CO2 (acid) in combustion and in cellular respiration, changing the color of bromothymol blue to yellow.
Learn more about pH in: brainly.com/question/491373
<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
