Question

50.0 mL of 0.275 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

Ka for HNO2 is 4.0x10^-4

Answer 1

pH formula's for salt's weak acidt strong base ->salft to pH = Ź [pkw + pka + 10g c] strong acid & weakbase salf toto pH = ₃ [ pkW _pkke_loge

Kaz 4x104 Ang molarity 20 - 275 | pka, log ka = log ve x104) volume 50ml=0.052 1 pka_ 3.398 M = 2 no of moles 0.27500-05 l pkw =14 = 0.01375 at equivalence point moles of moles of 0.01375 ANO - Nagh - Molarity of moles of Naolt Nach volume of Naolt 0-0 1325 volume of Nash volume of waoH = 0.013752 NaNo to salt ANO + NaOH weak acid strong base 1o.01375 0.81375 to 01375 Hv 1-0.01375 1-0.01375 o 10.01375 0.01375 -0.215686 M= IC] Total volume = 0.05+0.01375= 0.06375 [Nano 2] = 3 2 = -2 For this salt pH formula pH = + [ pkw tpka toge pH = Į [ 14 +3.398 + log (0.215686)] pH 8.366 Rateplesse

ANO: molasity 20. 275 ka= 4x104 1 pka_ -log ka = log (24x154) volume som = 0.054 pka_ 3.398 M= A no of moles= 0.27580-05 / pk w 214 20.01375 at equivalence point moles of moks of = 0.01375 a ANO - Nauh. Molarity of moles of Naolt Nach volume of Naolt 0.01375 volume of Nach volume of waoH = 0.013752 Aldaten ANO + NaOH weak acid strong base Nang + to salt n w220 o 01375 <-0.01.375 -0.01375 +0.01375 ET 0 0 10.01375 Total volume 0.057 0.01375 = 0.06375 0.01375 ĽNaNO3 = 3 :0.215686 M32C) r 0.06375 For this salt pH formula pH = + ( pkw tpka toge] pH = { [14+3.398 +log (0.215686)] pH = 8.366 Rateplease