1.trộn 100 ml dung dịch HCl 0,4M với 100 ml dung dịch Ca(OH)2 0,3M tính pH?
2.trộn 50ml dd Ba(OH)2 0,1M với 150ml dd H2SO4 0,04M
(Ba(OH)2, H2SO4 điện li hoan

Question

1.trộn 100 ml dung dịch HCl 0,4M với 100 ml dung dịch Ca(OH)2 0,3M tính pH? 
2.trộn 50ml dd Ba(OH)2 0,1M với 150ml dd H2SO4 0,04M
(Ba(OH)2, H2SO4 điện li hoan toàn theo 2 nấc)

diemloan · Answer

a/ nNaOH=0,2.0,1=0,02(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nNaOH=0,2.0,1=0,02(mol)nNaOH=0,2.0,1=0,02(mol)
NaOH&#x2192;Na++OH&#x2212;" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">NaOH→Na++OH−NaOH→Na++OH−
nNa+=nOH&#x2212;=0,02(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nNa+=nOH−=0,02(mol)nNa+=nOH−=0,02(mol)
&#x21D2;CMNa+=0,020,4+0,1=0,04(mol/l)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒CMNa+=0,020,4+0,1=0,04(mol/l)⇒CMNa+=0,020,4+0,1=0,04(mol/l)
nBa(OH)2=0,3.0,4=0,12(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nBa(OH)2=0,3.0,4=0,12(mol)nBa(OH)2=0,3.0,4=0,12(mol)
Ba(OH)2=Ba2++2OH&#x2212;" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">Ba(OH)2=Ba2++2OH−Ba(OH)2=Ba2++2OH−
&#x21D2;nOH&#x2212;=0,24(mol);nBa2+=0,12(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒nOH−=0,24(mol);nBa2+=0,12(mol)⇒nOH−=0,24(mol);nBa2+=0,12(mol)
&#x21D2;CMBa2+=0,120,5=0,24(mol/l)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒CMBa2+=0,120,5=0,24(mol/l)⇒CMBa2+=0,120,5=0,24(mol/l)
nOH&#x2212;=0,02+0,24=0,26(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nOH−=0,02+0,24=0,26(mol)nOH−=0,02+0,24=0,26(mol)
&#x21D2;CMOH&#x2212;=0,260,5=0,52(mol/l)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒CMOH−=0,260,5=0,52(mol/l)⇒CMOH−=0,260,5=0,52(mol/l)
b/ nHCl=0,2V(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nHCl=0,2V(mol)nHCl=0,2V(mol)
&#x21D2;nH+=nCl&#x2212;=0,2V" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒nH+=nCl−=0,2V⇒nH+=nCl−=0,2V
&#x21D2;CMCl&#x2212;=0,2V2V=0,1(mol/l)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒CMCl−=0,2V2V=0,1(mol/l)⇒CMCl−=0,2V2V=0,1(mol/l)
nH2SO4=0,3V(mol)=nH+2=nSO42&#x2212;" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nH2SO4=0,3V(mol)=nH+2=nSO2−4nH2SO4=0,3V(mol)=nH+2=nSO42−
&#x21D2;CMSO42&#x2212;=0,3V2V=0,15(mol/l)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒CMSO2−4=0,3V2V=0,15(mol/l)⇒CMSO42−=0,3V2V=0,15(mol/l)
nH+=0,2V+0,6V=0,8V(mol)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">nH+=0,2V+0,6V=0,8V(mol)nH+=0,2V+0,6V=0,8V(mol)
&#x21D2;CMH+=0,8V2V=0,4(mol/l)" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" tabindex="0">⇒CMH+=0,8V2V=0,4(mol/l)

manhtuan0608 · Answer

Giải thích các bước giải:
Bài 1:
$\begin{array}{l} n_{H^+}=n_{HCl}=0,4\times 0,1=0,04\ mol.\ n_{OH^-}=2n_{Ca(OH)_2}=0,3\times 0,1=0,03\ mol.\ \text{Vì}\ 0,04>0,03\ ⇒n_{H^+}\ \text{dư.}\ ⇒n_{H^+}\text{(dư)}=0,04-0,03=0,01\ mol.\ ⇒[H^+]=\dfrac{0,01}{0,1+0,1}=0,05\ M.\ ⇒pH=-\log([H^+])=-\log(0,05)=1,3\end{array}$
Bài 2:
\(\begin{array}{l} n_{OH^-}=2n_{Ba(OH)_2}=0,1\times 0,05=0,005\ mol.\ n_{H^+}=2n_{H_2SO_4}=2\times 0,15\times 0,04=0,012\ mol.\ \text{Vì}\ 0,005
chúc bạn học tốt!

Hoidap247 - Hỏi Đáp Bài Tập

1.trộn 100 ml dung dịch HCl 0,4M với 100 ml dung dịch Ca(OH)2 0,3M tính pH? 2.trộn 50ml dd Ba(OH)2 0,1M với 150ml dd H2SO4 0,04M (Ba(OH)2, H2SO4 điện li hoan toàn theo 2 nấc)

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1.trộn 100 ml dung dịch HCl 0,4M với 100 ml dung dịch Ca(OH)2 0,3M tính pH? 2.trộn 50ml dd Ba(OH)2 0,1M với 150ml dd H2SO4 0,04M (Ba(OH)2, H2SO4 điện li hoan toàn theo 2 nấc)

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