The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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Answer: to calculate pH use -log[H+] or - log[OH-]..the solution is basic as the “NaOH” is attached to a hydroxide.Since we need to find the pH (per hydrogen) and not the pOH( per hydroxide) we need to find the pOH of the substance first then we subtract that by 14 so we can arrive at the pH of the substance.
Explanation: So -log( 1 x 10^(-5)) = 5 which is the pOH.Now we subtract that by 14 which gives us -9 and now you’d multiply that by -1 bcuz we can’t have a negative so the pH of the substance is 9
