Question

A solution with weight percent of sodium hydroxide of 6% is prepared by adding the following mass of water to 200 grams of that solution with a weight percent of NaOH of 30%.Find the amount of water to be added?

Answer 1

Answer on the question #60484, Chemistry / Physical Chemistry

Question:

A solution with weight percent of sodium hydroxide of 6% is prepared by adding the following mass of water to 200 grams of that solution with a weight percent of NaOH of 30%. Find the amount of water to be added?

Solution:

The mass percent of NaOH in solution can be calculated as:

$$\omega = \frac {m (N a O H)}{m (N a O H) + m (H _ {2} O)},$$

where $m(NaOH)$ is the mass of sodium hydroxide and $m(H_2O)$ is the mass of water. $m(NaOH) + m(H_2O)$ is the sum of their masses, that means the mass of the solution.

Then, we can write a system of equations for former and resulting concentrations:

$$0.3 = \frac {m (N a O H)}{200},$$ $$0.06 = \frac {m (N a O H)}{200 + m ^ {\prime} (H _ {2} O)},$$

where $m^{\prime}(H_2O)$ is the mass of added water. If we divide the first equation by the last, we get:

$$\frac {0.3}{0.06} = \frac {200 + m ^ {\prime} (H _ {2} O)}{200},$$ $$5 * 200 = 200 + m ^ {\prime} (H _ {2} O)$$ $$m ^ {\prime} (H _ {2} O) = 800 \, g$$

Answer: 800g.

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