Which solution below would have the greatest buffering capacity? The solution was prepared from a weak acid and the salt of its conjugate base. 0.160 M C6H5OHCOOH and 0.160 M C6H5OHCOONa, Ka = 1.05e-3 0.0892 M H2NCH2COOH and 0.0892 M H2NCH2COOK, Ka = 4.50e-3 0.0725 M H2NCH2COOH and 0.0725 M H2NCH2COONa, Ka = 4.50e-3 0.1360 M C6H5OHCOOH and 0.1360 M C6H5OHCOOK, Ka = 1.05e-3

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

Explanation:

The first mixture

0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,

Ka = 1.05x 10⁻³ .

The second mixture

0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,

Ka = 4.5 x 10⁻³ .

The third mixture

.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,

Ka = 4.5 x 10⁻³

fourth mixture

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .

Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .

Ka is less if

Ka = 1.05 x 10⁻³ .

Dilute acids have greater buffering capacity

So ultimate answer is

0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,

Ka = 1.05x 10⁻³ .

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying