pH 5 6 heating NaClO converts the MnSO4 solution to MnO2 and the ion equation is

It should be the first one, but it's not that simple, please see the explanation.

1) Does the mnO2 in react with H+?

2) Why is OH- written in the environment PH 5 6

Manganese dioxide is generally stable, insoluble in water, dilute acids and bases, and does not disproportionate in acids and bases.

However, manganese dioxide is an amphoteric oxide (but the acidity and alkalinity are very low), and it can react with concentrated acid and concentrated alkali.

Mno2 and NaOH can produce Na2Mno3 (sodium manganate). Reaction with concentrated acid yields tetravalent manganese ions, but they are very unstable (strongly oxidizing) and are immediately converted to other forms (e.g., mno2 reacts with concentrated hydrochloric acid at low temperatures to form water and mncl4, which then decomposes into mncl2 and cl2).

It should be noted that Naclo is a strong alkaline substance (Naclo solution, pH above 10, the concentration of hydroxide ions cannot be ignored), so hydroxide ions can appear on the left side of the equation. (e.g., adjust HCl solution to pH 7 with NaOH and the reaction equation is HCH ===H2O). However, a pH of 5 to 6 is the final state, and it would be wrong to use oh ] in this case.

Under the condition of pH 5 6, MN element is the most stable substance with mno2 or mno(oh)2 (called manganese hydroxide oxide or hydrated manganese dioxide), and hypochlorous acid mainly exists in molecular form [at pH =, more than 95% of Clo- is converted to HCl, and at pH, more than 99% of ClO- is converted to HCl]. In other words, there will not be a lot of hydrogen ions in the final system, because it will be bound away by hypochlorite. Moreover, at a pH of 5 6, the concentration of hydrogen ions is less than 10-5mol L, which is chemically considered completely depleted, and this species is not suitable as a product to appear at the right end of the equation Mn2+

clo- +2oh- ===mno2 + cl- +h2o

In water, manganese dioxide is generally dissolved only in the presence of substances such as hydrogen peroxide, adipic acid, or ferrous sulfate.

For example: mNO2+H2O2+H2SO4 - >mnSO4+O2|+h2o

mno2+feso4+h2so4——>mnso4+fe2(so4)3+h2o

The acidic environment is, of course, the former equation.

2NaClO3+4HCl(concentrated)=2ClO2+Cl2+2H2O+2NaCl

The following method is better because "such HCl is more concentrated", but the essence is the same (the ion equation is the same).

naclo3+nacl+h2so4=clo2+1/2cl2+na2so4+h2o

The redox reaction occurs under the alkaline condition to generate hydroxide and sodium sulfate, and the masking ion equation is: 3H2O+3Cl - 6Fe 2+ =2Fe(OH) 3 +4Fe 3+ +3Cl -

So the answer is: 3H2O+3Clo - 6Fe 2+ =2Fe(OH) 3 +4Fe 3+ +3Cl -

1) Sodium bicarbonate ionizes sodium ions and bicarbonate ions in water, and the ionization equation is NaHCO3 NA+HCO3 - so the answer is: NaHCO3 NA+HCO3-

2) NaHSO4 is a strong electrolyte, NaHSO4 ionizes sodium ions and sulfate ions and hydrogen ions in water, NAHSO4NA+H+SO42-, so the answer is: NAHSO4NA+H + SO42-

h+ +oh-= h2o

Because the sodium bisulfate is completely ionized, in fact, the reaction before the rollback should be the reaction of alkali neutralization of large excitoic acid salts, and there is no change before and after the reaction of Na and SO4 ions, in fact, the reaction is just like the above-mentioned ion galena disorder.