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USPshnik [31]
3 years ago
9

Calculate the Ka for the following acid. Determine if it is a strong or weak acid. HClO2(aq) dissolves in aqueous solution to fo

rm H+(aq) and ClO2−(aq). At equilibrium, the concentrations of each of the species are as follows: [HClO2]=0.24M [H+]=0.051M [ClO2−]=0.051M
Chemistry
1 answer:
Talja [164]3 years ago
3 0

Answer:

The value of Ka = 1.1*10^{-2}

It is a weak  acid

Explanation:

   From the question we are told that

             The concentration of [HClO_2]=0.24M

             The concentration of  [H^+]=0.051M

             The concentration of  [ClO_2^-]=0.051M

Generally the equation for the ionic dissociation of HClO_2 is

                HClO_2_(aq) -------> H^{+}_{(aq)} + ClO_2^{-}_{(aq)}

The equilibrium constant is mathematically represented as

                         Ka = \frac{concentration  \ of  \ product  }{concentration \ of \  reactant }

                               = \frac{[H^+][ClO_2^-]}{[HClO_2]}

Substituting values since all value of concentration are at equilibrium

                    Ka = \frac{0.051 * 0.051}{0.24}

                          = 1.1*10^{-2}

Since the value of  is less than 1 it show that in water it dose not completely

disassociated  so it an acid that is weak

                         

               

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In Chapter 1 of Astrobiology, A Very Short Introduction, the author talks about some properties of one element in particular tha
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15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

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