Calculate the [OH - ] of 119.56 mL of a buffer initially consisting of 0.1234 M
ID: 962269 • Letter: C
Question
Calculate the [OH-] of 119.56 mL of a buffer initially consisting of 0.1234 M H2NCH2COOH and 0.0985 M H2NCH2COONa after addition of 0.0026 mol of NaOH. Assume that no volume change occurs after addition of the base. The Ka of H2NCH2COOH is 4.50e-3
2.613e-12
1.184e-12
3.826e-3
1.774e-12
please show working
Calculate the [OH-] of 119.56 mL of a buffer initially consisting of 0.1234 M H2NCH2COOH and 0.0985 M H2NCH2COONa after addition of 0.0026 mol of NaOH. Assume that no volume change occurs after addition of the base. The Ka of H2NCH2COOH is 4.50e-3
2.613e-12
1.184e-12
3.826e-3
1.774e-12
Explanation / Answer
when 0.0026 moles of NaOH is added, H2NCH2COOH reacts
H2NCH2COOH+NaOH----> H2NCH2COONa +H2O
this increases moles of H2NCH2COONa and decreases moles of H2NCH2COOH.
moles of H2NCH2COOH in 119.56ml of 0.1234M= 0.1234*119.56/1000=0.015
moles of NaOH= 0.0026 moles, moles of NH2CH2COOH remaining= 0.015-0.0026=0.0124
moles of NH2CH2COONa= initial moles+ moles formed during reaction= 0.0985*119.56/1000+0.0026 =0.0117+0.0026 =0.0143 moles
Concentrations : NH2CH2COOH= 0.0124/0.11956=0.1037M NH2CH2COONa= 0.0143/0.11956=0.119
pH= pKa+ log {[NH2CH2COONa]/ [NH2CH2COOH]}
pKa= -log(4.5/1000)= 2.346, pH= 2.346+log [0.119/0.1037) =2.41
pOH= 14-pH= 14-2.41=11.59, [OH-] =10(-11.59)= 2.57*10-12 ( A is the correct answer)
PH= 2.346+log(0.119