Question

A solution of two acids contains 3% hydrochloric acid. How much of the other acid should be added to 20 mL of the solution to reduce the
content of hydrochloric acid to 1%?

Answer 1

Answer:

hope it helps

Step-by-step explanation:

Expected HCl levels are about 20%. ... But there is a small amount of iron left in the suspension. ... Add 0.1 ml of hydrochloric acid to 210 ml of Hoxinine HCL in 0.5% solution to pH = 4.38.

Answer 2

Answer:

The answer is 40 ml  

Step-by-step explanation:

Given that a solution of two acids contains 3% of Hydrochloric acid (Hcl)

Then the percentage of other acid must be 97%

volume of the solution = 20 ml

Here we need to find how much of other acid must be add to reduce the content of hydrochloric acid to 1%

Given that 20 ml of solution contains 3% Hcl acid  

then content of Hcl in 20 ml solution = $\frac{3}{100} (20) = \frac{3}{5} = 0.6$ ml

Therefore, content of other acid = 20 ml - 0.6 ml = 19.4 ml

Now assume that x ml of other acid is added to the solution to reduce Hcl acid up to 1%. If concentration of Hcl is 1% then the concentration of other acid must be 99%

After adding x ml of other acid

volume of the solution = (20+x)ml  

the content of other acid in solution = (19.4+x)ml  

Therefore, concentration of other acid = $\frac{(19.4+X)}{(20+X)} (100) = 99$  

                                     ⇒       $(19.4+X)(100) = 99(20+X)$

                                     ⇒       1940 +100 x = 1980 + 99 x

                                     ⇒        100x - 99x = 1980 - 1940

                                     ⇒             x = 40 ml        

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