2n Ca(OH)2= n HCl
2[c Ca(OH)2 v Ca(OH)2] = c HCl v HCl
c Ca(OH)2= c HCl v HCl/2v Ca(OH)2
c Ca(OH)2= (0.158M)(32.40mL)/2(27mL)
c Ca(OH)2= 0.111M
pls. check if it's right:)
If 27.0 mL of Ca(OH)2 with an unknown concentration is neutralized by 32.40 mL of 0.185 M HCl, what is the concentration of the Ca(OH)2 solution?
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