Question

pH POH Label pH (calculated from molarity) pH (pHydrion) Solution, Concentration Acids HCI, 0.1 M CH3COOH, 0.1 M [H ] (calculated from meter) (calculated) (meter reading) 1.81 A1 13 B1 2.65 A2 HCI, 0.01 M CH3COOH, 0.01 M 2.24 2.78 B2 A3 HCI, 0.001 M 11 2.60 2.92 B3 pH CH3COOH, 0.001 M Solution, Concentration Bases NaOH, 0.1 M pH Label POH (calculated from molarity) (H+ (calculated from meter) (calculated) pH (pHydrion) (meter reading) 12.88 13 13 D1 11.07 C2 NH4OH, 0.1 M NaOH, 0.01 M NH4OH, 0.01 M 11.63 D2 10.23 C3 NaOH, 0.001 M 10.45 D3 NH4OH, 0.001 M. 9.71

*calculated from mete* column please

Answer 1

We know that pH = -log [t] and port - log [or] Thus. Here. We calculate [ht] from giver pt data Al label. pH meter reading 2 1.81 VID Similarly or [H] = 10-pH = 10-18 = 0.01551M) By label pH meter reading • 2.65 pH = -log[H] . [H] = 10 PH -2.65 =10 = 0.0022 This the table is given below pH PoH [HA] (Calculated meter PH Calculated Label Solution ) from meter) | p Hydrion concentration from molarity calculated reading) Acids AI HCl,01 M 13 1.81 10.0155m 2 BI CH₃COOH, o.1M| 12.65 0.0022 M 2 A2 Hel, 0.01M2 12 2.24 0.0058M 2 2.78 B2 CH₂ cooh, o OIM 0.0017 3 , 0.001M 3 11 2.60 0.0025M 4. 2.92 0.0012 M solution 14 pohlculated PH concentrations from molarity Calculated reading P meter [t] (Calculated pH Bases from meter) (p Hydrion) B3 CH3200 H,0.001M Klabel Solution Le NaOH, O.IM IDINHAOH, O.IM 162 Nao H₂0.01M 0.01M 12 2 12 12.88 11.07 | 11.63 10.23 110.45 11.32 x 10-13 ml 8.51x10-12 mlie 2.34*10-12 M 5.89x10-1'm lio 3.55x105"M 102NHAOH, 0.01M 3 pooh, 0.001M/3 9.71 1.9501010 13/NAGOH, 0.001m