balanced equation =
3Cu(OH)2 + 2H3PO4 → Cu3(PO4)2 + 6H2O
I know CaCO3 will form a precipitate because all carbonates do that. so your answer is B
Potassium hydroxide is a strong base and hydrobromic acid is a strong acid. This implies that the pH of the end-point [neutralization] of their titration will be around pH 7. A good indicator for this kind of pH is bromthymol blue. This is because this indicator changes its colour at pH 7.
PH + pOH = 14
11.8 + pOH = 14
pOH = 14 - 11.8
pOH = 2.2
[OH-] = 10 ^- pOH
[OH-] = 10 ^- 2.2
[OH-] = <span>6.33 x 10^-3 M
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Answer B
hope this helps!