Suppose a gas starts with a volume of 4.52L, temperature of 23 C, and pressure of 102,00 Pa. If the volume changes to 4.83L and a temperature of -12C what is the new temperature of the gas

The answer to your question is     P2 = 84.16 kPa

Explanation:

Data

Volume 1 = V1 = 4.52 L            Volume 2 = V2 = 4.83 l

Pressure 1 = P1 = 102 kPa        Pressure 2 = P2 = ?

Temperature 1 = T1 = 23°C      Temperature 2 = T2 = -12°C

Process

1.- Convert the temperature to °K

Temperature 1 = 23 + 273 = 296°K

Temperature 2 = -12 + 273 = 261°K

2.- Use the Combined Gas law to solve this problem

                  P1V1/T1 = P2V2/T2

-Solve for P2

                  P2 = P1V1T2 / T1V2

-Substitution

                  P2 = (102)(4.52)(261) / (296)(4.83)

-Simplification

                  P2 = 120331.44 / 1429.68

-Result

                  P2 = 84.16 kPa

the ratio of the conjugate base to the acid required to attain the desired ph.

what is the question here. i think you messed up.