The net ionic equation for NaOH and Na2SO4 when they form a precipitate is simple. It will contain only the atoms that participate in the reaction. Both of these compounds are soluble.
2 NaHSO4 + 2 H2O ----> Na2SO4 + 2 H3O+ + SO4-2
NaOH + KHT = KOH + NaHT
2NaOH + H2SO4 = Na2SO4 + 2H2O Ionic Equation 2Na + 2OH + 2H + SO4 = 2Na + SO4 + 2H2O Cross out common elements and compounds on both sides to get the ionic equation: 2OH + 2H = 2H2O
The precipitate is copper(II) hydroxide. The chemical reaction is:2 NaOH + CuSO4 = Cu(OH)2 + Na2SO4
Na+OH- + Cu+2CL-2 -> Na+Cl- + Cu+2(OH)-2
This reaction is:2 NaOH + H2SO4 = Na2SO4 + 2 H2O
H+ + OH- = H2O
CuSO4 +2 NaOH = Cu (OH) 2 ↓ + Na2SO4 Cu (OH) 2 blue flocculent precipitate
sulfuric acid + sodium hydroxide ---> sodium sulfate + water
Balanced equation. 2NaOH + H2SO4 -> Na2SO4 + 2H2O 12.5 grams NaOH (1 mole NaOH/39.998 grams)(1 mole Na2SO4/2 mole NaOH)(142.05 grams/1 mole Na2SO4) = 22.2 grams sodium sulfate produced ===========================
H+ + oh- ---> h2o
Hf + oh- ---> f- + h2o
The chemical equation is:3 NaOH + FeBr3 = 3 NaBr + Fe(OH)3