Answer:
1x10^-8 M
Explanation:
Since the solution turns blue, it mean the solution is a base.
Now, to know which option is correct, we need to determine the pH of each solution. This is illustrated below:
1. Concentration of Hydrogen ion, [H+] = 1x10^-2 M
pH =..?
pH = - log [H+]
pH = - log 1x10^-2
pH = 2
2. Concentration of Hydrogen ion, [H+] = 5x10-2 M
pH =..?
pH = - log [H+]
pH = - log 5x10^-2
pH = 1.3
3. Concentration of Hydrogen ion, [H+] = 5x10 M
pH =..?
pH = - log [H+]
pH = - log 5x10
pH = - 1.7
4. Concentration of Hydrogen ion, [H+] = 1x10-8 M
pH =..?
pH = - log [H+]
pH = - log 1x10^-8
pH = 8
A pH reading shows if the solution is acidic or basic. A pH reading between 0 and 6 indicates an acidic solution, a pH reading of 7 indicates a neutral solution while a pH reading between 8 and 14 indicates a basic solution.
From the above calculations, the pH reading indicates a basic solution when the hydrogen ion concentration was 1x10^-8 M.
Answer:
3
Explanation:
Applying,
= R/R'............... Equation 1
Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.
From the question,
Given: R = 40 atoms, R' = 5 atoms
Substitute these values into equation 1
= 40/5
= 8
= 2³
Equation the base,
n' = 3
Answer:
The new concentration is 0.125 M.
Explanation:
Given data:
Initial volume V₁ = 125.0 mL
Initial molarity M₁ = 0.150 M
New volume V₂ = 25 mL +125 mL = 150 mL
New concentration M₂ = ?
Solution:
M₁V₁ = M₂V₂
0.150 M × 125 mL = M₂ × 150 mL
M₂ = 0.150 M × 125 mL / 150mL
M₂ = 18.75 M.mL/150 mL
M₂ = 0.125 M
The new concentration is 0.125 M.
Answer:
they're losing electrolytes
Explanation:
When athletes sweat, they're losing electrolytes primarily in the form of sodium (Na+) and chloride (Cl-), so when you start to replace lost fluids, ahtletes should replace the electrolytes as well. Potassium (K+), Magnesium (Mg2+) and Calcium (Ca2+) are electrolytes also lost through sweating.
Answer:
pH = 12.61
Explanation:
First of all, we determine, the milimoles of base:
0.120 M = mmoles / 300 mL
mmoles = 300 mL . 0120 M = 36 mmoles
Now, we determine the milimoles of acid:
0.200 M = mmoles / 100 mL
mmoles = 100 mL . 0.200M = 20 mmoles
This is the neutralization:
HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O
20 mmol 36 mmol 20 mmol
16 mmol
We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:
NaHCOO → Na⁺ + HCOO⁻
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹
These contribution of OH⁻ to the solution is insignificant because the Kb is very small
So: [OH⁻] = 16 mmol / 400 mL → 0.04 M
- log [OH⁻] = pOH → 1.39
pH = 14 - pOH → 12.61
