Question

18x10 and 10 Min Ho see sample esse 172 Calculate the formate concentration for 'l and pH of a solution that is 0.50 M informacion

please show work and use the ICE box to solve

please use this picture instead, I posted the wrong problem.

thank u!

the bottom picture is the right one

18x10 and 10 Min Ho see sample esse 172 Calculate the formate concentration for 'l and pH of a solution that is 0.50 M informacion

here is another picture, hopefully it is clearer

Answer 1

Ansa + HCOOH is a weak acid & HNO3 is a strong acid. Ka q HCOOH = 1.8 x 104 . HCOH Capi) - Ht Cap.) + H (co-cap:) ſu coon) = 0.05, (HNO3) = 011M => H Cao u (opo) ut lap.) + ucco-Capi) Sitial (07? 0:05M 01 0 Charge xH tx M + RM At epm (105-x) M Coʻl tu) M ka= [4+]{1600 ] 1 [ucan] Kaz 1.88 104 = a (olta) +x M 9.1 Swall 0.os-x 1.9 x 16 y = x (0.1) 0905 x = 0.05 x logx to'xlo 0,1 X=0,09p X10-3 lomp . = 9 x16s 1 [ucat] =r=9x105ml (x is small. Hence, I assumption is true pH = - Lag [+] (x is small] (4+3=ooltx =011 pus - Lay (0-1) = 1 CS Scanned with CamScanner TpH = 1.

Ausu Cang Canlas) - Ht (ap) + Couscos Cap-) . Acid o Conjugate Base Cols Con is a acid because it is proten donar Collsca- is conjugate base of acid, Cs Scanned with Hs cool CamScanner

- Henders in Haselbalsch epn is pu - pkat Log i salt Acid? pH=4 [Acid}= Benzoic acid concentrations oudM ned wiškanned with Sodium benze eite - nelmScannis alt} = concentration of

- pka= 4:2 [of densic aud] auds Ka= binric u => 4 = 4.2 + log [scit] 6:2 -0.2= Lag [salt] - 0.2 01 63 = [salt] [salt}= 0.126M [cots Leonat] = 0.131 => sodium benzoate 2013 M. concentration of CamScanner