Respuesta :
the Given the initial ratio of the sulfuric acid to water
[tex]\begin{gathered} \text{Solution 1}\Rightarrow\text{ 3 : 7 } \\ \text{Solution 2}\Rightarrow\text{ 4 : 9} \end{gathered}[/tex]So to obtain the ratio value of the sulfuric in solution 1, multiply each water ratio by 3/7
[tex]\begin{gathered} 7\Rightarrow\text{ 7}\times\frac{3}{7}=\frac{21}{7}\text{ = 3} \\ 21\Rightarrow\text{ 21}\times\frac{3}{7}=3\times3=6 \\ 42\Rightarrow\text{ 42}\times\frac{3}{7}=6\times3=18 \\ 63\Rightarrow63\times\frac{3}{7}=9\times3=27 \end{gathered}[/tex]
So to obtain the ratio value of the sulfuric in solution 2, multiply each water ratio by 4/9
[tex]\begin{gathered} 9\Rightarrow9\times\frac{4}{9}=1\times4=4 \\ 18\Rightarrow18\times\frac{4}{9}=2\times4=8 \\ 45\Rightarrow45\times\frac{4}{9}=5\times4=20 \\ 63\Rightarrow63\times\frac{4}{9}=7\times4=28 \end{gathered}[/tex]In conclusion, you will notice that when solution 1 has a water ratio of 63, the acidic content is 27, while in solution 2 when water ratio is 63 the acidic content is 28.
Hence, Solution 2 is more acidic
