<em>Calculate the pH of the following substances formed during a volcanic eruption:
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<em>• Acid rain if the [H +] is 1.9 x 10-5
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<em>• Sulfurous acid if [H +] = 0.10
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<em>• Nitric acid if [H +] = 0.11</em>
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<h3>Further explanation </h3>
pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.
pH = - log [H⁺]
![\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1.9%5Ctimes%2010%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D5-log1.9%5C%5C%5C%5CpH%3D4.72)
![\tt pH=-log[10^{-1}]\\\\pH=1](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B10%5E%7B-1%7D%5D%5C%5C%5C%5CpH%3D1)
![\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B11%5Ctimes%2010%5E%7B-2%7D%5D%5C%5C%5C%5CpH%3D2-log~11%3D0.959)
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
pH = 13.1
Explanation:
Hello there!
In this case, according to the given information, we can set up the following equation:
Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:
Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:
And the resulting concentration of KOH and OH ions as this is a strong base:
![[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5Cfrac%7B0.00576mol%7D%7B0.012L%2B0.032L%7D%3D0.131M)
And the resulting pH is:
Regards!
Answer:
If the concentration of product D is increased, the rate of the reverse reaction would increase.
Explanation:
Chemical reaction:
A + B ⇄ C + D
In given condition the equilibrium is disturb by increasing the concentration of product.
When the concentration of product D is increased the system will proceed in backward direction in order to regain the equilibrium. Because when the product concentration is high it means reaction is not on equilibrium state the reaction will proceed backward direction to regain the equilibrium state.
According to the Le- Chatelier principle,
At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.
The equilibrium can be disturb,
By changing the concentration
By changing the volume
By changing the pressure
By changing the temperature
Answer: D. the hardening of liquid magma that leads to igneous rock
Explanation:
