Question

1. a 1.00 l sample of hcl(aq) is prepared and has a ph of 1.045. what mass of hcl(g) is dissolved?

Explanation:

Step1: Find \([H^+]\) from pH

The formula for pH is \(pH = -\log_{10}[H^+]\). Rearranging to solve for \([H^+]\), we get \([H^+]=10^{-pH}\). Substituting \(pH = 1.045\), we have \([H^+]=10^{-1.045}\). Calculating this, \(10^{-1.045}\approx0.0903\space M\).

Step2: Determine moles of HCl

HCl is a strong acid, so it dissociates completely in water: \(HCl(aq)
ightarrow H^+(aq)+Cl^-(aq)\). This means the concentration of \(HCl\) is equal to the concentration of \(H^+\). For a 1.00 L solution, moles of \(HCl\), \(n = [HCl]\times V\). Since \([HCl]=[H^+]=0.0903\space M\) and \(V = 1.00\space L\), \(n = 0.0903\space mol\).

Step3: Calculate mass of HCl

The molar mass of \(HCl\) is \(M = 1.008\space g/mol + 35.45\space g/mol = 36.458\space g/mol\). Using the formula \(m = n\times M\), we substitute \(n = 0.0903\space mol\) and \(M = 36.458\space g/mol\). So \(m = 0.0903\space mol\times36.458\space g/mol\approx3.29\space g\).

Answer:

The mass of \(HCl(g)\) dissolved is approximately \(\boldsymbol{3.29\space g}\) (or more precisely, depending on significant figures, around 3.3 g or as calculated with more precision).