Answer:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:
Answer:
a: before equivalence point
b: equivalence point
c: before equivalence point
d: after the eqivalence point
e: before equivalence point
f: after the eqivalence point
Explanation:
Balanced equation of reaction:
NaOH +HCl =NaCl +H2O;
Volume of HCl is fixed and it 100ml and concentration is 1.0M
N1 and N2 normality of HCl and NaOH respectively;
V1 and V2 volume of HCl and NaOH respectively;
we have given molarity but we need normality;
<em>but in case of NaOH and HCl n-factor is 1 for each.</em>
hence
normality=molarity;
At equivalence point:
Before equivalence point :
After the equivalence point:
case a: 5.00 mL of 1.00 M NaOH
hence it is before equivalence point
case b: 100mL of 1.00 M NaOH
hence it is equivalence point
case c: 10.0 mL of 1.00 M NaOH
hence it is before equivalence point
case d: 150 mL of 1.00 M NaOH
hence it is after the eqivalence point
case e: 50.0 mL of 1.00 M NaOH
hence it is before equivalence point
case f: 200 mL of 1.00 M NaOH
hence it is after the eqivalence point