Answer:
pH = 7.1581
Explanation:
The equilibrium of NaHSO₃ with Na₂SO₃ is:
HSO₃⁻ ⇄ SO₃²⁻ + H⁺
Where K of equilibrium is the Ka2: 6.5x10⁻⁸
HSO₃⁺ reacts with NaOH, thus:
HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺
As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:
HSO₃⁻: 0.252 moles
SO₃²⁻: 0.139 moles
Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:
0.0500L ₓ (1mol /L): 0.050 moles of NaOH.
Thus, final moles of both compounds are:
HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles
SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles
Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:
pH = pka + log [SO₃²⁻] / [HSO₃⁻]
Where pKa is - log Ka = 7.187
Replacing:
pH = 7.187 + log [0.189] / [0.202]
Answer:
The pOH is 4, 44 and the solution is basic.
Explanation:
The pOH is a measure of the concentration of OH (hydroxyl) ions in the solution.
The pOH is calculated as :
pOH = -log (OH-)= -log (3.67 x 10-5 )= 4, 44. In this case, the solution is basic.
The p0H scale ranges from values of 0 to 14 (less than 7.0 is basic and greater than 7.0 is acidic, a pH = 7.0 is neutral)
Answer:
1.7820210165667731 your welcome :)
Explanation:
Answer:
6.321 × 10^22
Explanation:
Mass of Oxygen =
3.36
g
Molar mass of oxygen (
O
2
) = 16 x 2 =
32
g
m
o
l
−
Total molecules in oxygen = Mass in grams/Molar mass x
N
A
=
3.36
32
x
6.02
x
10
23
=
6.321
x
10
22
Note:
N
A
(Avagadro's number) =
6.02
x
10
23
Hope it helps...
After 100years, sample is 250g
After 200 years, sample is 125g
After 300years, sample is 62.5 g
Answer:
0.190 M
Explanation:
Let's consider the neutralization reaction between HCl and NaOH.
HCl + NaOH = NaCl + H2O
11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:
0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol
The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.
1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:
M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M
Answer:
The initial concentration of HCl was 0.1904 M
Explanation:
Step 1: Data given
Volume of HCl solution = 10.0 mL = 0.010 L
Volume of a NaOH solution = 11.9 mL = 0.0119 L
Molarity of NaOH solution = 0.160 M
Step 2: The balanced equation
HCl + NaOH → NaCl + H2O
Step 3: Calculate the concentration of HCl
C1*V1 = C2*V2
⇒with C1 = the concentration HCl = TO BE DETERMINED
⇒with V1 = the volume of HCl = 0.010 L
⇒with C2 = the concentration of NaOH = 0.160 M
⇒with V2 = the volume of NaOH = 0.0119 L
C1 * 0.010 L = 0.160 M * 0.0119 L
C1 = (0.160 M * 0.0119 L) / 0.010 L
C1 = 0.1904 M
The initial concentration of HCl was 0.1904 M