Propionic acid (CH3CH-COOH) has a K, 1.3 x 10^-5 and phosphoric acid (H3PO2) has a Ka = 7.5 x 10^-3 Choose the conjugate base for each.
A. CH3CH2COO2- for CH3CH2COOH; HPO4 2- for H3PO4
B. CH3CH2CO- for CH3CH2COOH; H2PO3 - for H3PO4
C. CH3CH2COOH2 for CH3CH2COOH; H4PO4 for H3PO4
D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4

D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4.

Explanation:

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In this case, since the ionization of a weak acid produces a conjugate acid, molecule to which the leaving H⁺ is added to, and a conjugate base, everything that remains after the withdrawal of the H⁺, we can note down the ionization of each acid down below:

Thus, we notice that the answer is D. CH3CH2COO- for CH3CH2COOH; H2PO4 - for H3PO4 because the phosphoric acid is most likely to ionize to dihydrogen phosphate ions rather than phosphate, or hydrogen phosphate ions it undergoes a stepwise ionization in which the first step is the predominant one.

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when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml