Question

100.0 ml of hcl required 25.00 ml of 0.2000 m mg(oh)₂ to reach ph 7. what is the concentration of the hcl? the neutralization equation is 2 hcl + mg(oh)₂ → mgcl₂ + 2 h₂o there is enough information to calculate the moles of mg(oh)₂ 1 mol mg(oh)₂ ( 2 l mg(oh)₂ (----------------) = 3 mol mg(oh)₂ 1 l then use stoichiometry to convert moles of mg(oh)₂ into moles of hcl 4 mol hcl ( 5 mol mg(oh)₂)(----------------) = 6 mol hcl 7 mol mg(oh)₂ we now can calculate the concentration of hcl since we have both the moles of hcl and the volume of hcl 8 mol hcl concentration = ------------ = 9 mol/l 10 l hcl a. acid b. base c. hcl d. naoh e. nacl f. h₂o g. 25.00 h. 0.02500 i. 0.2000 j. 1.000 k. 0.005000 l. 100.0 m. 0.1000 n. 0.05000 o. 2.000 p. 0.01000 q. 0.001000 r. 58.31 s. 5.831

Explanation:

Step1: Calculate moles of \(Mg(OH)_2\)

Use the formula \(n = C\times V\). The volume \(V\) of \(Mg(OH)_2\) is \(25.00\space mL=0.02500\space L\) (conversion: \(1\space L = 1000\space mL\)), and concentration \(C = 0.2000\space M\).
\(n_{Mg(OH)_2}=0.02500\space L\times0.2000\space mol/L = 0.005000\space mol\)

Step2: Convert moles of \(Mg(OH)_2\) to moles of \(HCl\)

From the balanced equation \(2HCl+Mg(OH)_2
ightarrow MgCl_2 + 2H_2O\), the mole ratio of \(HCl\) to \(Mg(OH)_2\) is \(2:1\).
If \(n_{Mg(OH)_2}=0.005000\space mol\), then \(n_{HCl}=0.005000\space mol\times\frac{2\space mol\space HCl}{1\space mol\space Mg(OH)_2}=0.01000\space mol\)

Step3: Calculate concentration of \(HCl\)

The volume of \(HCl\) is \(V = 100.0\space mL = 0.1000\space L\). Use the formula \(C=\frac{n}{V}\)
\(C_{HCl}=\frac{0.01000\space mol}{0.1000\space L}=0.1000\space mol/L\)

Answer:

1. K. \(0.005000\)
2. H. \(0.02500\)
3. K. \(0.005000\)
4. P. \(0.01000\)
5. K. \(0.005000\)
6. P. \(0.01000\)
7. J. \(1.000\)
8. P. \(0.01000\)
9. M. \(0.1000\)
10. L. \(100.0\)