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# Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

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We are given the reaction as: - - - - > Response To) The pH will be 12.36,We have to convert the concentrations of HCl and NaOH in moles,So we have, n(HCl) = (0.0290) L X (0.290 mol/L) = 8.41 X moles and for NaOH, we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13 X moles Now, it seems that the NaOH is in excess, so that the remaining amount will be; 1.13 X - 8.41 X = 2.89 X moles Now, the total volume will be as = 0.0390 + 0.0290 = 0.068 L Thus, the concentration of [] = 2.89×10ˉ3 mol / 0.068 L = 4.25 X M pOH = - log [] = -log (4.25 x) = 1.37 Therefore, pH = 14 - pOH = 14 - 1.37 = 12.6 So that the pH of the the solution will be 12.6, which is basic in nature. Answer B) The pH will be 1.68, Now, to the concentration that we need to find moles of The HCl and the NaOH also;n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X mol n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X mol here we can see, the HCl is in excess of the amount for the rest of will be; 8.41 X - 7.41 X = 1.0 X mol Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L The concentration of [HCl] = 1.0 X mol / 0.0480 L = 2.08 X M Which is = [H⁺] So, pH = - log [] = -log(2.08 X ) = 1.68 therefore, the pH will be 1.63, which is more acidic in nature.