A10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point?

A10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point? for hydrocyanic acid, pka = 9.31

Answers

When the titration of HCN with NaOH is:

HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume 

                      = 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

                                          = 2 mmol / 0.0998mL

                                          = 20 mL

and according to the molar ratio so, moles of CN- = 2 mmol

∴the molarity of CN- =  moles / total volume 

                                   = 2 mmol / (10mL + 20mL ) = 0.0662 M

when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

            = 14 - 9.31 = 4.69 

when Pkb = -㏒Kb

         4.69 = -㏒ Kb 

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq) 

by using the ICE table:

∴ the initials concentration are:

[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

[HCN] = [OH-]= X

when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114 

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

                    = -㏒ 0.00114

POH = 2.94

∴PH = 14 - 2.94 = 11.06

 

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. hbr completely dissociates to h^+ and br- ions. a is incorrect because h_2s does not fully ionize to h^+ and hs^- ions and some remain as the compound h_2s. c and d are incorrect because, li_2o and libr do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both li_2o and libr have a complete octet so can not accept any electrons which disqualifies them as acids.

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