Question
Consider the neutralization reaction:$$2 \mathrm{HCl}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{CaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$$What volume of $0.1 \mathrm{M} \mathrm{HCl}$ is required to completely neutralize a 35.0-mL sample of $0.285 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2} ?$
Step 1
We do this by dividing by 1000: $$ 35.0 \, \text{mL} \, \text{Ca(OH)}_{2} = \frac{35.0}{1000} \, \text{L} \, \text{Ca(OH)}_{2} = 0.035 \, \text{L} \, \text{Ca(OH)}_{2} $$ Show more…
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