Chemistry, 17.08.2019 19:10 brittanyfox411

A25.00-ml sample of propionic acid, hc3h5o2, of unknown concentration was titrated with 0.141 m koh. the equivalence point was reached when 43.76 ml of base had been added. what is the hydroxide-ion concentration at the equivalence point? ka for propionic acid is 1.3 × 10–5 at 25°c. a. 1.5 × 10-9 m b. 1.1 × 10-3 m c. 1.1 × 10-5 m d. 8.3 × 10-6 m e. 1.0 × 10-7 m

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Answer from: jeremytheprimerapper

Concentration of hydroxide-ion at equivalence point = $8.3\times 10^{-6}M$

Explanation:

$HC_{3}H_{5}O_{2}+KOH\rightarrow C_{3}H_{5}O_{2}^{-}K^{+}+H_{2}O$

1 mol of $HC_{3}H_{5}O_{2}$ reacts with 1 mol of KOH to produce 1 mol of $C_{3}H_{5}O_{2}^{-}$

At equivalence point, all $HC_{3}H_{5}O_{2}$ gets converted to $C_{3}H_{5}O_{2}^{-}$ .

Moles of $C_{3}H_{5}O_{2}^{-}$ produced at equivalence point is equal to moles of KOH added to reach equivalence point.

So, moles of $C_{3}H_{5}O_{2}^{-}$ produced = $\frac{43.76\times 0.141}{1000}moles=0.00617moles$

Total volume of solution at equivalence point = (25.00+43.76) mL = 68.76 mL

Concentration of $C_{3}H_{5}O_{2}^{-}$ at equivalence point = $\frac{0.00617\times 1000}{68.76}M=0.0897M$

$OH^{-}$ produced at equivalence point is due to hydrolysis of $C_{3}H_{5}O_{2}^{-}$ . We have to construct an ICE table to calculate concentration of $OH^{-}$ at equivalence point.

$C_{3}H_{5}O_{}^{-}+H_{2}O\rightleftharpoons HC_{3}H_{5}O_{2}+OH^{-}$

I:0.0897 0 0

C: -x +x +x

E: 0.0897-x x x

$\frac{[HC_{3}H_{5}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]}=K_{b}(C_{3}H_{5}O_{2}^{-})=\frac{10^{-14}}{K_{a}(HC_{3}H_{5}O_{2})}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.0897-x}=7.69\times 10^{-10}$

or, $x^{2}+(7.69\times 10^{-10}\times x)-(6.90\times 10^{-11})=0$

So, $x=\frac{-(7.69\times 10^{-10})+\sqrt{(7.69\times 10^{-10})^{2}+(4\times 6.90\times 10^{-11})}}{2}$ M = $8.3\times 10^{-6}M$

So, concentration of hydroxide-ion at equivalence point = x M = $8.3\times 10^{-6}M$

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1