moles Ca(OH)2 = M x L = 0.2 x 1.0 = 0.2 mole.
moles HCl = M x L = 0.1 x 2.5 = 0.25 mole.
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
This is a limiting reagent problem and you need to determine which is the limiting reagent. The pH will be determined by the OTHER reagent.
Convert moles Ca(OH)2 to moles CaCl2.
Convert moles HCl to moles CaCl2.
The two answers will be different (and only one can be correct). The SMALLER one is ALWAYS the correct value in limiting reagent problems and the reagent producing the smaller value will be the limiting reagent.
Now convert the limiting reagent moles to moles of the non-limiting reagent, subtract from the original value of moles, and the excess will determine the (H^+) or (OH)2 (as the case may be) and pH = -log(H^+). Post your work if you get stuck. The above sounds like it can be confusing; however, if you follow step by step things work out very well.
If 1.0 L of 0.2 M Ca(OH)2 is added to 2.5 L of 0.1 M HCl, the pH of the final solution is?
can someone show me the working out of this question? im very confused by it. thankyou
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1 answer