Question

Exercise 14.47 Review Part Four solutions of unknown HCl concentration are titrated with solutions of NaOH. The following table lists the volume of each unknown HCI solution, the volume of NaOH solution required to reach the equivalence point, and the concentration of each NaOH solution. Calculate the concentration (in M) of the unknown HCl solution in the third case. HCI Volume (ml) NaOH Volume (mL) (NaOH (M) 24.00 mL 13.88 mL 0.1178 M Express your answer using four significant figures. HCI Volume (mL) 20.00 mL 14.00 mL 24.00 mL 6.00 mL NaOH Volume (ml) 27.44 mL 21.22 mL 13.88 mL 7.88 ml (NaOH (M) 0.1191 M 0.0332 M 0.1178 M 0.1285 M 90 AED ROQ? Concentration Submit Request Answer

Answer 1

THEL + Noul am) all from above eaction, Naon is lil Inac tlH₂O face (1) mol tatio between HCl and - NHIL -1 NHCl 1. Naohi = Equivalence Nooth point we know, men .n) moles 1. mo) molarity U2) Volume =) (n = mu] utvolume [HI] Uitel = [Nowot] Nash - 2 given [haw] =? Cncold) = 0.1178m HCL = 24.com Nooy = 13.88mi CHC) = [Nowh] Noot - Co. 1118m) (13.88 m2) - Vita - C24100D "0.06813m [4-significant figure :And T0.068131 m Hue to deach Equivalence point.