今欲用H3PO4与NaOH来配制pH = 7.20的缓冲溶液,则H3PO4与NaOH物质的量之比 n(H3PO4)∶n(NaOH)应当是 ( )今欲用H3PO4与NaOH来配制pH = 7.20的缓冲溶液,则H3PO4与NaOH物质的量之比 n(H3PO4)∶n(NaOH)应当是 ( ) (H3PO4的

今欲用H3PO4与NaOH来配制pH = 7.20的缓冲溶液,则H3PO4与NaOH物质的量之比 n(H3PO4)∶n(NaOH)应当是 ( )今欲用H3PO4与NaOH来配制pH = 7.20的缓冲溶液,则H3PO4与NaOH物质的量之比 n(H3PO4)∶n(NaOH)应当是 ( ) (H3PO4的 今欲用H3PO4与Na2HPO4来配制pH = 7.2的缓冲溶液,则H3PO4与Na2HPO4物质的量之比n(H3PO4)∶n(Na2HPO4)应当是（）?(H3PO4的pKa1~pKa3分别是2.12,7.20,12.36) 3,为什么? 今欲用H3PO4与Na2HPO4来配制pH = 7.2的缓冲溶液,则H3PO4与Na2HPO4物质的量之比n(H3PO4)∶n(Na2HPO4)应当(H3PO4的pKa1~pKa3分别是2.12,7.20,12.36) 3,为什么? 用0.025mol/LH3PO4和0.025mol/LNAOH来配制的PH值为12.0的缓冲溶液100ML,试计算所需H3PO4和NAOH的体积,已知PKA1=2.17,PKA2=7.20,PKA3=12.32 .今欲配制一pH=7.20的缓冲溶液,所用0.10 mol/L H3PO4和0.10 mol/L NaOH溶液的体积比是（ ）.今欲配制一pH=7.20的缓冲溶液,所用0.10 mol/L H3PO4和0.10 mol/L NaOH溶液的体积比是（ ）（H3PO4：pKa1=2.16,pKa2=7.21,pKa3= 由0.1mol/l H3PO4与0.1molNAOH配制ph=2.16缓冲溶液,则其体积比为?已知H3PO4的PKa=2.16 .今欲配制一pH=7.20的缓冲溶液,所用0.10 mol/L H3PO4和0.10 mol/L NaOH溶液的体积比是（ ）（H3PO4：pKa1=2.16,pKa2=7.21,pKa3=12.32） H3Po4与NaOH可以配制缓冲溶液吗?为什么?要满足什么要求? 配制500ML,PH=7.21的缓冲溶液,问需要0.10MOL`L-1NAH2PO4和0.20MOL`L-1NAOH各多少ML?（H3PO4：PKA 2=7.2 10 mL 0.20 mol·L-1H3PO4与15 mL 0.20 mol·L-1NaOH溶液混合后,求溶液的pH(H3PO4的Ka1 =10-2.12,Ka2 =10-7.20,Ka 3=10-12. pH=8.93KH2PO4-NaOH缓冲液怎么配制 今用H3PO4和Na3PO4配制PH=7.2的缓冲溶液H3PO4与Na3PO4的物质的量之比是（H3PO4的PKa1,PKa2,PKa3分别是2.12,7.20,12.4）A,一比二 B二比三 C三比二 D,一比一 C(H3PO4)=0.1mol/L与C(NaOH)=0.25mol/l等体积混合后,ph=?(H3PO4:Ka1=7.5*10-3,Ka2=6.2*10-8ka3=2,2*10-13PH=PKa2+lgNaH2po4/Na2Hpo4,这样算的答案与正确答案不符啊,是方法不对吗?麻烦下喽 求算pH的问题100mL,0.01mol/L H3PO4 与 100ml,0.02mol/L NaOH 溶液混合,求此时pH 急问大学分析化学：用NaOH溶液滴定H3PO4溶液至pH = 4.7时,溶液的简化质子条件为 ((H3PO4的pKa1~pKa3分别是2.12、7.20、12.36) (A) [H3PO4] = [H2PO4-] (B) [H2PO4-] = [HPO42-] (C) [H3PO4] = [HPO42-] (D) [H3PO4] = 2[PO43-] 选 在1L、0.1 mol·L-1H3PO4溶液中,加入6 g NaOH固体[M(NaOH)=40.0],完全溶解后,求溶液的pH在1L、0.1 mol·L-1H3PO4溶液中,加入6 g NaOH固体[M(NaOH)=40.0],完全溶解后,求溶液的pH.假设加入NaOH固体后溶液体积不变.（2 怎样来配制PH=2.0的PBS 溶液? 配制pH=7.21的缓冲溶液500ml,0.10MOL.NAH2PO4和0.20MOL.LNAOH各多少ML?（H3PO4:PKA2=7.21)加急,麻烦帮忙做下,