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Question: What quantity (in mL) of a \(45\% \) acid solution of a mono-protic strong acid must be mixed with a...

What quantity (in mL) of a 45%45\% acid solution of a mono-protic strong acid must be mixed with a 20%20\% solution of the same acid to produce 800  mL{\rm{800}}\;{\rm{mL}} of a 29.875%29.875\% acid solution?
A. 316316
B. 330330
C. 320320
D. 325325

Explanation

Solution

We know that the mixture is produced by adding solute into solvent. Here the volume of one acid is mixed with the same acid with different concentrations to produce total percent of acidic solution. The question demands the answer should be in mL unit.

Complete answer
In the given question if we convert the percentage value to volume then it will be easy to find the quantity in mL units.
Assume that V be the volume of 45%45\% acid solution.
We have 800  mL{\rm{800}}\;{\rm{mL}} as the final volume hence, the volume of 20%20\% acid solution will be (800  mLV)\left( {800\;{\rm{mL}} - {\rm{V}}} \right). Equating the amount of acids before and after mixing, we will have;
V×(45100)  +  (800mL    V)×(20100)  =  (800  mL)×(29.875100)V \times \left( {\dfrac{{45}}{{100}}} \right)\; + \;\left( {800\,mL\; - \;V} \right)\times \left( {\dfrac{{20}}{{100}}} \right)\; = \;\left( {800\;mL} \right)\times \left( {\dfrac{{29.875}}{{100}}} \right)
Solving the above equation, we can get value of V;

45V + \left( {800\;mL - V} \right)\left( {20} \right) = \left( {800\;mL} \right)\left( {29.875} \right)\\\ 45V - 20V + 16000\;mL = 23900\;mL\\\ 25V + 16000\;mL = 23900\;mL$$

V = \dfrac{{23900;mL - 16000;mL}}{{25}}\\
= 316;mL

**Hence, the answer is (A). Note: ** 1\. To find the volume it is very important to convert the percentage in terms of volume. 2\. Mono-protic acids are the acids which are having one transferable hydrogen ion (proton). For example, HCl or hydrochloric acid. 3\. The concentration is in percentage that means if there is ${\rm{100}}\;{\rm{mL}}$ solvent in which x ml acid is present, that means the percentage of acid is x.