The balanced equation for the reaction between H₂SO₄ and NaOH is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
stoichiometric point is also called the equivalence point. this is the point at which equal quantities of acid and base are added and all the H⁺ ions and OH⁻ ions react with each other.
the number of NaOH moles reacted - molar concentration x volume
number of NaOH moles = 0.3223 mol/L x 0.04556 L = 0.01468 mol
according to stoichiometry
2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.01468 mol of NaOH reacts with - 1/2 x 0.01468 mol = 0.007340 mol
the number of H₂SO₄ moles in 59.34 mL - 0.007340 mol
Therefore number of H₂SO₄ moles in 1000 mL - 0.007340 mol / 59.34 x 1000 = 0.1237 mol
molar concentration of H₂SO₄ is 0.1237 M
Here N H 4 O H { NH }_{ 4 }OH NH4OH is not a strong electrolyte because it doesn't dissociates completely.
Answer:
[C₆H₅NH₃⁺] = 0.0399 M
Explanation:
This excersise can be easily solved by the Henderson Hasselbach equation
C₆H₅NH₃Cl → C₆H₅NH₃⁺ + Cl⁻
pOH = pKb + log (salt/base)
As we have value of pH, we need to determine the pOH
14 - pH = pOH
pOH = 8.43 (14 - 5.57)
Now we replace data:
pOH = pKb + log ( C₆H₅NH₃⁺/ C₆H₅NH₂ )
8.43 = 9.13 + log ( C₆H₅NH₃⁺ / 0.2 )
-0.7 = log ( C₆H₅NH₃⁺ / 0.2 )
10⁻⁰'⁷ = C₆H₅NH₃⁺ / 0.2
0.19952 = C₆H₅NH₃⁺ / 0.2
C₆H₅NH₃⁺ = 0.19952 . 0.2 = 0.0399 M
Answer:
Average amu is 6.52556
Explanation:
Average the numbers
Add the two masses together.
6.01512+ 7.01600= 13.03112
Then divide by 2, we divide by 2 because that is how many #'s we are given.
13.03112/ 2= 6.52556
