Answer to Question #107414 in General Chemistry for kim

Question

Like all equilibrium constants, the value of Kw depends on temperature. At body temperature (37∘C), Kw=2.4⋅10^−14

1) What is the [H3O+] in pure water at body temperature?
2) What is pH of pure water at body temperature?

What mass of HClO4 should be present in 0.600 L of solution to obtain a solution with each pH value?

1) pH = 2.70
2) pH = 1.50
3) pH = 0.60

Answer 1

1. $K_w=2.4*10^{−14}=[H^+][OH^-]=[H^+]^2 \implies pH=pK_w/2=6.562$
2. pH of pure water at room temperature is 7. It is considered to be neutral.

$HClO_4$ is considered to be a strong acid, thus it dissociates completely. pH of strong acid is calculated using the formula :

$pH=-ln[H^+]=-ln[HClO_4]=-lnc$

$\implies c=e^{-pH}$ ---(b)

where c= concentration of $HClO_4 = H^+$concentration

Now, $c=Molarity of HClO_4=mass(HClO_4)/(99.5*0.6)$ , since molecular weight of the compound is 99.5 grams.

$\implies c=m/59.7$ ---(a)

Combining (a) and (b) we get;

$m=59.7e^{-pH}$

1) $pH=2.7 \implies m=59.7e^{-2.7}=4.012gms.$

2) $pH=1.5 \implies m=59.7e^{-1.5}=13.321gms.$

3) $pH=0.6 \implies m=59.7e^{-0.6}=32.764gms.$