Question

I was able to kind of figure out #24, but if someone could explain it to me so that it makes sense and do #25 that would be great!

24. Gangplank filled a beaker with 12.35 mL of 20 mM HCl. Gangplank slowly added 5.6 mL of 20 mM sodium hydroxide to the beaker. HCl and NaOH are a strong acid and strong base, respectively. An acid base reaction occurred in the beaker. What are the concentrations of the hydroxide ion, and the hydrogen ion after the reaction has gone to completion? mol m= 0-02m HCIX U35L= 2,5 x109 mul HCl m.L- mul 1.4 *JU4 mul HICI 0.62 m NGOHX 6.asoL11101 NaUH Co.012754U.0USOL) 1. 1x 16^ moi NauH x Imol HC) -1xl0 nul HCl veacad reactont -4 2.5x1U x10 ' mul HCI inirially -1110nul HC)-1,4 Y , 25. Balance the equation below, and identify which species was oxidized, and which was reduced. FeCl26) + Sn) → SnCl4s) + Fe()

Answer 1

Solution:

24. The total volume of reagents added = 12.35 + 5.6 = 17.95 mL

       NaOH + HCl $\rightarrow$ NaCl + H2O

As per the above reaction,

one mole of HCl will react with one mole of NaOH.

Both NaOH and HCl have the same concentration of 20 mM.

Therefore 5.6 mL of 20mM NaOH will neutralise 5.6 mL of 20 mM HCl.

There will be an excess of HCl equivalent to (12.35 -5.6 =) 6.75 mL of 20 mM HCl

By definition of molarity, 1000 mL of 20 mM HCl = 20 x 10^-3 moles.

Therefore 6.75 mL of 20 mM HCl = 20 x 10^-3 x 6.75 / 1000 = 0.135 x 10^-3 moles.

This 0.135 x 10^-3 moles of HCl present in the final volume of 17.95 mL

For molarity we have to calculate the number of moles in 1 L or 1000 mL.

Therefore molarity of HCl in the resultant solution = 0.135 x 10^-3 x 1000/17.95 mol/L

       or the concentration of HCl in the resultant solution = 7.52 x 10^-3 mol / L or M

HCl + H₂O $\rightarrow$ H₃O⁺ + Cl^-   one mole of HCl releases 1 mole of H⁺ ions in water.

Therefore 7,52 x 10^-3 M HCl = 7.52 x 10^-3 M of H⁺

Thus the resultant solution has the concentration of H⁺ ion = 7.52 x 10^-3 M

P^H = 2.124 Therefore P^OH = 14 - 2.124 = 11.876   [ since P^H + P^OH = 14]

- log [OH] = 11.876

log [OH] = -11.876 , Therefore [OH] = 1.33 x 10^-12 M

The OH^- ion concentration in the resultant solution is = 1.33 x 10^-12 M

25. FeCl₂ + Sn $\rightarrow$ SnCl₄ + Fe

     Multiple FeCl₂ by 2 to balance Cl. There will be a need to balance Fe by multiplying Fe by 2

Thus

2 FeCl₂ + Sn $\rightarrow$ SnCl₄ + 2Fe ( Balanced )

On the LHS, Fe in FeCl₂ is in +2 oxidation state. It reduces to elemental iron on the RHS

On the LHS Sn is in the elemental state oxidised to +4 oxidation state in SnCl₄ on the RHS.

Thus Sn is oxidised and Fe is reduced.