Question- Share with your friends Share 0 Vartika Jain answered this Dear Student, Moles of NaOH=Moles of [OH-] =0.1M ×0.05 L = 0.005 molMoles of H3PO4 = 0.15×0.06=0.009 molNow, H3PO4 ↔ H+ + H2PO4- 0.009 0 00.009-x x xK1=[H+][H2PO4-][H3PO4]10-3=(x)(x)(0.009-x)as x is very small, 0.009-x≈0.0090.009×10-3=x2or, x2= 9×10-6or, x= 3×10-3Therefore, [H+]=[H2PO4-]=0.003and, H2PO4- ↔ H+ + HPO42- 0.003 0 00.003-x x xK2=[H+][HPO42-][H2PO4-]10-8=(x)(x)(0.003-x)as x is very small, 0.003-x≈0.0030.003×10-8=x2or, x2= 30×10-12or, x= 5.47×10-6Therefore, [H+]=[HPO42-]=5.47×10-6K3 is very small and will yield very low H+ and so can be neglected. Now, total moles of H+from acid= 0.003+0.00000547 =0.00300547Since moles of OH- is more than H+, therefore the pH will be around 6-7. Hence, correct answer is (D) -1 View Full Answer
