The strong acid HCl was titrated with 1.00M NaOH. The HCl solution had a volume

Question

The strong acid HCl was titrated with 1.00M NaOH. The HCl solution had a volume of 100mL and a concentration of 0.100M. Find the pH at the following volumes of added base (Vb). Show the calculations (Hint: figure out how many moles of H3O+ (or OH-) are present at each point and divide by the volume of the solution at that point (you will have to figure it out) and use the [H3O+] to calculate pH): a) Vb = 0mL b) Vb = 1mL c) Vb = 5mL d) Vb = 9mL e) Vb = 9.9mL f) Vb = 9.99mL g) Vb = 10mL h) Vb = 10.01mL i) Vb = 10.1mL j) Vb = 12mL k) Vb = 15mL l) Vb = 20mL Please show all work, thank you!! If there is a way to use an ICE table, please solve that way, as that is how I understand best; if not any detailed explanation is greatly appreciated!

Explanation / Answer

a)Concentration of HCl= 0.1M, Volume =100ml =0.1L

when no NaOH is added, Concentration of HCl=0.1M

HCl +H2O----> H3O+ +Cl-

HCl being strong acid, ionizes completely

Hence [H3O+] =0.1M and pH= -log [H3O+]= 1

The given volumes of NaOH suggest in all the experiments, NaOH is limiting reactant

b) Mole of HCl in 100ml =Molairty* Volume (L) 0.1*0.1= 0.01moles

Moles of NaOH in 1ml of 0.1M= 0.1*1/1000=0.0001 moles

The reaction is HCl +NaOH---> NaCl +H2O

mole of HCl reacts with 1 mole of NaOH

NaOH is limting reactant and moles of HCl left after reaction =0.01-0.0001=0.0099

Volume after mixing = 100+1= 101ml =0.101 L Concentration of HCl =0.0099/ 0.101=0.09802M

pH= -log (0.09802) =1.0008

c) moles of NaOH in 5ml =0.1*5/1000 =0.0005, moles of HCl remaining after reaction=0.01-0.0005=0.0095

Volume after mixing = 100+5=105ml =0.105L, concentration of HCl now =0.0095/0.105=0.094, pH= 1.034

d) moles of NaOH in 9ml =0.0009, moles of HCl remaining = 0.01-0.0009=0.0091, Volume after mixing =100+9=109ml=0.109L, Concentration of HCl =0.0091/0.109=0.0084, pH= 1.078

e) when 9.9ml of NaOH is added = 9.9*0.1/1000=0.00099, moles of HCl remaining =0.01-0.00099= 0.00901, Volume after mixing = 100+9.99= 109.99ml= 0.10999 Concentration of HCl =0.00901/0.10999 =0.0819, pH= 1.086

f) when 9.99ml of NaOH is added =9.99*0.1/1000=0.000999, HCl remaining = 0.01-0.000999=0.009001 moles and Volume after mixing = 100+9.99= 109.99ml =0.10999 L, Concentration of HCl = 0.00901/ 0.10999=0.08192, pH= 1.0863

g) moles in 10ml NaOH= 10*0.1/1000=0.001, moles of HCl remaining = 0.01-0.001=0.009, Volume after mixing = 100+10=110ml=0.11L , Concentration of HCl after mixing = 0.009/0.11=0.0818, pH= 1.087

h) Moles of 10.01ml NaOH= 10.01*0.1/1000=0.001001, moles of HCl remaining =0.01-0.001001=0.008999, Volume after mixing = 100+10.01= 110.01ml= 0.11001L, Concentration of HCl = 0.008999/0.11001=0.081802 M pH= 1.0872

i)moles in 10.1ml of NaOH= 10.1*0.1/1000=0.00101, moles of HCl remainiing=0.01-0.00101= 0.00899, Volume after mixign = 10.1+100= 110.1ml=0.1101L, Concentration after mixing = 0.00899/0.1101=0.0816M, pH=1.088

j) moles in 12ml of NaOH =12*0.1/1000=0.012, moles of HCl remaining =0.01-0.012= 0.0088 , Volume after mixnig =112ml =0.112L, Concentrationp = 0.0088/0.112 = 0.079, pH= 1.104

k) Moles in 15ml NaOH= 15*0.1/1000= 0.0015, moles o HCl remaining =0.01-0.0015=0.0085, Volume after mixing = 100+15= 115ml=0.115L, Concentration of HCl =0.0085/0.115=0.07391, pH=1.13

l) moles in 20ml NaOH= 20*0.1/1000=0.002, moles of HCl remaining =0.01-0.002=0.008, Volume after mixing =100+20=120ml=0.12L, Concentration = 0.008/0.12=0.067, pH=1.176

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