a)Calculate the pH when 6.80 g of NaBrO is added to 100. mL ofa 0.500M HBr0 solu

Question

a)Calculate the pH when 6.80 g of NaBrO is added to 100. mL ofa 0.500M HBr0 solution (assume no change in volume). pko 8T 8.76 Nabro b) Calculate the pH when 100 mL of 0.50 M NaOH is added to 100 mL of a 0.50M HBrO solution. Nacrt8ro- Na BrottHio ol c) Calculate the pH when 100 mL of 1.0 M NaOH is added to 100 mL of a 0.50M HBrO solution. 13.4 The acidity of soil is determine by mixing a sample of soil with water and measuring the pH of the resulting slurry. Scientists have found that mixing the soil into 0.10M CaCh instead of water provides more consistent results. Would you expect the pH of the slurry with CaCl2

Explanation / Answer

Answer:

(b) Given volume of NaOH =100 mL=0.1 L, (since 1L=1000 mL) molarity=0.5 M=0.5 mol/L

Therefore moles of NaOH=molarity x volume=0.5 mol/L x 0.1 L=0.05 mol.

HBrO, volume=100 ml=0.1 L, molarity=0.5 M=0.5 mol/L

Therefore moles of HBrO=0.5 mol/L x 0.1 L=0.05 mol.

The neutralization reaction between NaOH and HBrO is

NaOH + HBrO --------> NaBrO + H2O

The mole ratio between NaOH and HBrO is 1:1.

Here also the mole ratio between NaOH:HBrO=0.05 :0.05=1:1

There this is equivalence point. At this point only species of BrO^- is present which is conjugate base of weak acid HBrO.

Given pKa=8.71 for HBrO,

Ka=10^-8.71=1.949 x10^-9. (-logKa=pKa=8.71)

Kb=Kw/Ka=(1x10^-14)/(1.949x10^-9)

Kb=5.128x10^-6.

So BrO^- react with water to make basic solution.

Total volume=100+100 ml=200 mL=0.2 L.

[BrO^-]=0.05 mol/0.2 L=0.25 M.

Now ICE table,

BrO^- + H2O <--------> HBrO + OH^-

Initial. 0.25 . 0 . 0

Change. -x . +x . + x

Equilibrium .0.25-x. x. x

Ka=[HBrO][OH^-]/[BrO^-]

5.128 x 10^-6= (x^2)/(0.25-x)

Since Ka<<<<x, then (0.25-x=0.25)

Therefore x^2=0.25 x 5.128 x10^-6

x=(1.282 x 10^-6)^(1/2)

x=0.001132=[OH^-]

pOH=-log(0.001132)=2.95

pH=14-pOH=14-2.95=11.

Answer pH=11.

(c) Given volume of HBrO=100 ml=0.1 LL and molarity=0.5 M=0.5 mol/L.

Moles of HBrO=0.5 mol/L x 0.1 L=0.05 mol.

Volume of NaOH=100 mL=0.1 L, molarity=1 M=1 mol/L

Moles of NaOH=1 mol/L x 0.1 L=0.1 mol

According to neutralization reaction (see in part b), here excess of NaOH is present.

So excess moles of OH^- =0.1 mol - 0.05 mol=0.05 mol.

Total volume=100+100=200 ml=0.2 L,

Molarity of [OH^-]=0.05 mol/0.2 L=0.25

pOH= - log(0.25)=0.6.

pH=14-pOH=14-0.6=13.4.

pH=13.4.

Please let me know if you have any doubt. Thanks and I hope you like it.

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