Question

show work when solving the table please, thank you

pH of pure water 7,00 130* 10/10-"M(OH) .0X10-”M KW 1.0210-14 Initial Burette Reading 0.00 mL (KOH) 6,100M pH KOH Additions [H30+] * [OH-] + Kw Min = M2 V 2 5.01010-12 1.96810 3 9.427x65 6.1M) (1) = m2 (51) 12.51*10-12 3.85X10319.66176-15 11.86x10-12 19927710 vs bo 101190 Burette Reading 1.00 mL [ 11.30 2.0 mL 11.60 3.00mL 1173 4.00 mL 5.00 ml 11.98 6.00 mL 7.00 ml 4.00 ml 12.15 9.10mLL 12.11 10.00m 12:21 (0.1)(w 50100) 17.04 12.08 Augo KW Show a complete worked out set of calculations for the third addition

Answer 1

KOH addetions Κω Burette Reading pH HOT [OHT 1.00mL 11.30 5.01810 1.96810 9.88 1015 2.00mL 11.60 2.518 1612 3.98×10 9.94 10-15 3.00 mL 11.73 1.8681072 5.37810 9.988107 -4:00 mL 11.85 1.4181012 7.07 8103 9.9681015 500 mL 11-98 1.048 1012 9.5x10 9.8881015 600m 12-04 9:12x1013 10.96X163 9.99 xrots 700m 1208 8.31x1013 12.02x10 ² 9.988105 8.00mL 12.13 7.41 x 1013 13.480 100 9.98810ls 9.00 mL 12.17 .7h81013 14.798103 9.99x10ls 10.00 m 12.21 6.16*10-13 16.2% 103 10.781015 алчлар 100-16 XID kw - 10 . -15 * ph= -log[43 ot] [113077² (105 by * pon = -log [on] [ou] - 10rpon = 10.01810 s 1x10 * PH + pol = 14