Hello, please give me advice from my high school chemistry teacher

When using graphite electrode to electrolyze copper sulfate solution, hydroxide ions should be used for the anode electrode reaction, but water should be written when writing the total reaction. Water is a weak electrolyte, but in a normal aqueous solution, water can ionize a small amount of hydroxide, and this small amount of hydroxide will be discharged, that is, it will lose electrons, and eventually become water and oxygen. The hydroxide of water ionization is consumed, and according to the equilibrium principle, when the concentration of the product decreases, the equilibrium shifts in the direction of positive reaction, so the ionization of water is promoted.

In this way, the hydroxide in the water is continuously discharged at the anode, and the hydrogen ions ionized by the water accumulate at the anode, which eventually makes the solution acidic, that is, the sulfuric acid solution is finally generated.

In the same way, in the sodium hydroxide solution, the hydrogen ions ionized by water are discharged at the cathode, resulting in hydrogen gas being released, leaving hydroxide, so the pH near the cathode is greater than 7, and the electrode reaction formula should be written as hydrogen ions; At the anode, the hydroxide root that is ionized by water is discharged, and eventually it becomes oxygen and water, leaving hydrogen ions, so the pH near the anode is less than 7, and the electrode reaction formula should be written as hydroxide. In the whole process, only hydroxide and hydrogen ions are discharged, and the amount of sodium hydroxide is not reduced, so the electrolysis of sodium hydroxide, which is equivalent to the electrolysis of water, should be written as the reaction equation of electrolysis of water when writing the total reaction. As the water is electrolyzed, the concentration of the solution increases, so eventually the pH of the whole solution increases!

The chemical equilibrium constant is to be united, and for different reactions, the unit of the equilibrium constant is different, which can be determined according to the expression of the equilibrium constant.

I don't know if I explained it this way, did you understand?

The above problem anode is represented by hydroxide ions,,, cathode reaction water in sodium hydroxide solution, and the water will ionize to produce hydrogen ions. The unit of chemical equilibrium constant should be determined by the chemical reaction. Chemistry covers a wide range of topics, and students should figure it out on their own.

Chemistry students float by... It's easy to say, but you just look for a teacher

d。Precipitate Mg2 with NaOH and turn HCO3 into CO32; then use BaCl2 to precipitate SO42- and CO32; then use Na2CO3 to precipitate Ca2 and excess Ba2+; Finally, HCL was used to remove Sun Qing's liquid to remove CO32- and OH-.

b。According to the principle of charge balancing, positive and negative charges *2 x =

a 。The n atoms in the 5 ammonium positive friends are oxidized, and 3 of the 5 nitrate atoms are reduced, so the ratio of the number of nitrogen atoms that are oxidized to the number of nitrogen atoms that are reduced is 5 3.

1.Because the amount added each time is not easy to determine, only an excessive amount of substances can be added, so in order to remove SO42- it is necessary to add a sufficient amount of Ba2+, so that after removing sulfate, new barium ion impurities will be produced, in order to remove it, a sufficient amount of sodium carbonate must be added, so there is a new carbonate impurity, in order to remove it, hydrochloric acid must be added until it happens not to be in the bubble, as for other impurities in the process of removing barium ions, It is worth mentioning that magnesium ions and carbonate will be double hydrolyzed to produce magnesium hydroxide precipitation, but you haven't learned this yet, so you specially use sodium hydroxide to remove magnesium ions, and put it in the first step, choose D

2.According to the conservation of charge, it is easy to obtain the sum of the positive charges of the cation: 2* should be equal to the sum of the negative charges of the anion: calculate that x= is the amount of sulfate substance, and choose b

3.From the equation, it can be seen that the original product of oxidation and also making trouble is nitrogen, which is 0 valence, so the ammonium pants closed root ion should rise 3 valence is oxidized, nitrate should be reduced by 5 valence, two nitrate groups are not valence, so there are three nitrate atoms in the nitrate are reduced, and all five root ions are oxidized, so it is 5:3 to choose a

5.Using the conservation of charge, positive charge = negative charge , so x is solved as.

The NO3- in it is the NO3- in NH4NO3, so the redox Qingbian reaction is 5 NH4+ and 3 NO3- in the nitrogen of the oak, which generates N2, so there are 5 N atoms that are oxidized by the spring and 3 that are reduced, so the ratio is 5:3, choose A

The addition of reagent is an excess of Tansui, Na2CO3 can remove the excess of BaCl2 and HCl can remove the excess of Nikuku CO32-

Mg2 is 1mol in MgCl2, Cl is 1mol, Cl is 1mol in K2So4, and K is 1mol in So42-

Assuming that the volume is x liters, then the cl- in CaCl2 and the Cl- in AlCl3 are added together.

c = The pH change after mixing CaCl2 with AlCl3 should not be considered here, resulting in the hydrolysis of Al3+ to precipitate).