Question

4 points Determine the pH of a solution prepared by mixing 60.0 mL of 0.025 M HCl with 40.0 mL of 0.035 M HBr. answer in Question 7. I leo Mathull ditarit

Answer 1

Volume of HC1 = 60.0 mL molarity of HC1 = 0.025 M moles of HC1 = molarity of Hax volume of Ha 1 x60.0 mL X-pol mi 1L - X 60 = 0.025 Mx mol/L M = 0.0015 mol Volume of HBr = 40.0 mL molarity of HBr = 0.035 M moles of HBr = molarity of HBr x volume of HBr U = 0.035 M x mol/L x 40.0 mL 1000 mL M *40.0 mL = 0.0014 mol Moles of H+ = moles of HBr +moles of HCI = 0.0014 mol +0.0015 mol = 0.0029 mol Total volume = Volume of HCI+ Volume of HBr = 60.0 mL+40.0 mL = 100.0 mL molarity is defined as the number of moles of solute present is 1 L volume of solution. molarity, M = moles of solute Volume of solution concentration of H' = Total volume _Moles of H+ 0.0029 mol 1000 mL 100.0 mL 1L (H* )= 0.029 mol/L [H*]= 0.029 M - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Calculate the pH as follows: pH = -log[*] = -log(0.029) = 2.54 Therefore, the required solution is 2.54