Answer:
At standard room temperature, ![[{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20OH%5E%7B-%7D%5D%20%5Capprox%202.5%20%5Ctimes%2010%5E%7B-7%7D%5C%3B%20%5Crm%20M)
when ![[{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%5E%7B%2B%7D%5D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%3B%20%5Crm%20M)
.
Explanation:
The following equilibrium goes on in water:
.
The forward reaction is known as the self-ionization of water. The ionization constant of water, 
, gives the equilibrium position of this reaction:
![K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}]](https://tex.z-dn.net/?f=K_%7B%5Crm%20w%7D%20%3D%20%5B%7B%5Crm%20H%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20OH%5E%7B-%7D%7D%5D)
.
At standard room temperature (
), 
. Also, ![[{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%5E%7B%2B%7D%7D%5D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
. Substitute both values into the equation and solve for ![[{\rm OH^{-}}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20OH%5E%7B-%7D%7D%5D)
.
![\begin{aligned} {[}{\rm OH^{-}}{]} &= \frac{K_{\rm w}}{[{\rm H^{+}}]} \\ &\approx \frac{10^{-14}}{4.0 \times 10^{-8}} = 2.5 \times 10^{-7}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%7B%5B%7D%7B%5Crm%20OH%5E%7B-%7D%7D%7B%5D%7D%20%26%3D%20%5Cfrac%7BK_%7B%5Crm%20w%7D%7D%7B%5B%7B%5Crm%20H%5E%7B%2B%7D%7D%5D%7D%20%5C%5C%20%26%5Capprox%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B4.0%20%5Ctimes%2010%5E%7B-8%7D%7D%20%3D%202.5%20%5Ctimes%2010%5E%7B-7%7D%5Cend%7Baligned%7D)
.
In other words, in an aqueous solution at standard room temperature, when .
Probably the kinetic energy of rubbing the pump against the valve, combined with the increase in gas pressure in the tire itself.
Answer is: <span>A. 18.02 g/mol.
At standard temperature and pressure 1 mol of gas occupied 22.4 liters:
V(H</span>₂O) = 22.4 L; volume of water.
Vm = 22.4 L/mol; molar volume at STP.
n(H₂O) = V ÷ Vm.
n(H₂O) = 22.4 L ÷ 22.4 L/mol.
n(H₂O) = 1 mol; amount of substance (water).
M(H₂O) = Ar(O) + 2Ar(H) · g/mol.
M(H₂O) = 16 + 2 ·1.01 · g/mol.
M(H₂O) = 18.02 g/mol; molar mass of water.
