An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would

$($ Given $\log 2=0.30)$

<p>An aqueous solution of HCl initially has a pH of 1.0, meaning the hydrogen ion concentration is $[\text{H}^+]=10^{-1} \, \text{M}$.</p> <p>When we dilute the solution by adding an equal volume of water, the concentration of hydrogen ions will be halved:</p> <p>$ [\text{H}^+]_{\text{new}} = \frac{10^{-1}}{2} = 5 \times 10^{-2} \, \text{M} $</p> <p>To find the new pH, we use the pH formula:</p> <p>$ \text{pH} = -\log[\text{H}^+] $</p> <p>Substituting the new hydrogen ion concentration gives:</p> <p>$ \text{pH} = -\log(5 \times 10^{-2}) $</p> <p>Using the logarithm property $\log(a \times b) = \log a + \log b$:</p> <p>$ \text{pH} = -(\log 5 + \log 10^{-2}) = -\log 5 + 2 $</p> <p>Given $\log 2 = 0.30$, we need $\log 5$, which can be calculated as:</p> <p>$ \log 10 = \log(2 \times 5) = \log 2 + \log 5 \Rightarrow \log 5 = \log 10 - \log 2 = 1 - 0.30 = 0.70 $</p> <p>Thus:</p> <p>$ \text{pH} = -(0.70) + 2 = 1.3 $</p> <p>Therefore, the pH of the solution increases to 1.3.</p>