Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
C. Its oxidation number increases.
Explanation:
- <em><u>Oxidation is defined as the loss of electrons by an atom while reduction is the gain of electrons by an atom</u></em>.
- Atoms of elements have an oxidation number of Zero in their elemental state.
- When an atom looses electrons it undergoes oxidation and its oxidation number increases.
- For example, <em><u>an atom of sodium (Na) at its elemental state has an oxidation number of 0. When the sodium atom looses an electrons it becomes a cation, Na+, with an oxidation number of +1 , the loss of electron shows an increase in oxidation number from 0 to +1.</u></em>
D as the two substances could have close boiling points, so if the temperature is not controlled well, both substances will evaporate.
For an non spontaneous reaction between silver (Ag) and copper (Cu) and their ions, Cu is the oxidizing agent while Ag+ is the reducing agent,
The following reactions will take place;
Anode Cu = Cu+2 + 2e- E= +0.34 volts
Cathode; Ag+ + e = Ag E = +0.80 volts
The net reaction will be Cu + 2Ag+ = Cu+2 + 2Ag
Thus, the voltage will be
= +0.80 - (+0.34)
