Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
One way they are different is that they show elevation
The answer is B
I also did this quiz.
<span>The reason it will be 7 for some titrations is that when you titrates a strong acid with a strong base for example HCl and NaOH the salt formed is conjugate base of strong acid and will be a very weak base
That means that it cannot produce any OH^-1 and all the H+ has been converted to water.The only source of H+ or OH is water with a Ka of 10^-14 so the pH = -log [H+]=-log 10^-7 = 7
second reason is
When you titrates a weak acid with strong base at equivalence point
only a water solution of the conjugate base exists
CH3COOH + NaOH ----- Na+ CH3COO^-1 + H2O
Since the conjugate base is the conjugate base of a weak acid it will hydrolyze in water like so
for instance Na+ CH3COO^-1 + HCl---- CH3COOH + NaCl the equivalence point will be way BELOW 7 and in the case of above will be less than 5. So pH of 7 at equivalence point is only reached in strong acid strong base titrations.
hope this helps</span>
