Answer:
0.0095 moles of acid were neutralized by the antiacid
Explanation:
The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:
<em>Moles HCl added:</em>
42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl
<em>Moles NaOH to titrate the excess:</em>
10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.
<em>Moles of acid that were neutralized:</em>
0.0105 moles - 0.0010 moles =
<h3>0.0095 moles of acid were neutralized by the antiacid</h3>
Answer:
Concentration of Na₂SO₃ is 0.0533M
Explanation:
To obtain concentration of Na₂SO₃ it is necessary to obtain moles of primary standard, KIO₃ in the standarization:
0.4991g × (1mol / 214g) = 2.332x10⁻³ mol KIO₃ / 0.2500L = 9.329x10⁻³M
As you take 50.00mL of this solution, moles are:
0.05000L × (9.329x10⁻³mol / L) = <em>4.664x10⁻⁴ moles KIO₃</em>
Now, the reaction of Na₂SO₃ with KIO₃ is:
KIO₃(aq) + 3 Na₂SO₃(aq) → KI(aq) + 3 Na₂SO₄(aq)
That means 1 mole of KIO₃ reacts with 3 moles of Na₂SO₃. Thus, moles of Na₂SO₃ that reacts are:
4.664x10⁻⁴ moles KIO₃ × (3 mol Na₂SO₃ / 1 mol KIO₃) = <em>1.399x10⁻³ mol Na₂SO₃</em>
As 26.25mL of titrant were consumed, concentration of Na₂SO₃ is:
1.399x10⁻³ mol Na₂SO₃ / 0.02625L = <em>0.0533M</em>
My Friday is going good and I might get to go over to my BFF house!!!
Reactant at the left product at the rig