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saveliy_v [14]
3 years ago
6

Does the volume of a liquid change

Chemistry
1 answer:
nydimaria [60]3 years ago
6 0

Answer:

Yes it does change

Explanation:

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The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.
olchik [2.2K]

Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
  • V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L.

Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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