Question

eneral Chemistry 11-Karan Activities and Due Dates HW 15 o 5/22/2016 11:00 PM 0 75/100。5/22/2016 02:33 AM re Grade -Print Calculator 보 Periodic Table Question 7 of 22 Map presented by Calculate the pH of the resulting solution if 31.0 mL of 0.310 M HCI(aq) is added to (a) 41.0 mL of 0.310 M NaOH(aq). Number pH= (b) 21.0 mL of 0.410 M NaOH(aq). Number pH = 100

Answer 1

(a) HC1+ NaOH → NaCl + H2O Volume ofacid HCl 31.0mL - 31.0mL 1000mL 0.031L Concentration of HCl = 0.310M mol No ofmoles of HC-0.031L × 0.310- 9.61x 10-3 mol Volume ofNaOH added 41.0 mL 41.0mL× -0.041 0.310M 1000mL Concentration of NaOH mol No of moles of KOHadded 0.041L x 0.310_ 1.27x10-2 mol No of moles of base left out (1.27 x 10%-9.61x 10-3) mol 3.10x10-3 mol Total volume= (31.0+41.0) mL = 72.0mL = 0.0720L 3.10x10 mol -0.0720L Concentration of acid, "OH = 0.0431 M pOH =-log | -OH -log (0.0431) pH-14.0-1.37 12.6 Hence, pH after the addition of 41.0 mL base is 12.6

(b) HC1+ NaOH → NaCl + H2O Volume of acid HC1 31.0mL = 31.0mL 1000 mL 0.031L Concentration of HC1 0.310M mol Noofmoles ofHCl=0.031L × 0.310- = 9.61 x 10-3 mol Volume of NaOH added 21.0mL 21.0mL x 1000mL = 0.021L Concentration of NaOH= 0.410M mol No ofmoles of KOH added = 0.021 L × 0.410 8.61x10-3 mol No of moles of acid left out = (9.61x 10-3 mol-8.61 × 10-3 mol = 1.00x10" mol (31.0+21.0) mL 52.0 mL 0.0520し Total volume 1.00x10-3 mol 0.0520L Concentration of acid, OH 0.0192M log(0.0192) = 1 . 72 Hence, pH after the addition of 21.0 mL base is 1.72