pH= 11 => pOH=3

=> n_OH^- = 10^-3V₁

pH= 10 => pOH= 4

=> n_OH^- = 10^-4V₂

10^-3V₁=10^-4V₂

_=> V₂= 10V₁ => V₂ = 1000ml

V_H2O = V₂ - V₁ =900ml

\(pH=11\Rightarrow C_M=10^{-3}M;\\pH=10\Rightarrow C_M=10^{-4} \)

Coi rằng nước là dd NaOH C_M = 0 M

Ta có:

\(\Rightarrow V=\frac{100ml\cdot\left(10^{-3}-10^{-4}\right)}{\left(10^{-4}-0\right)}=900\left(ml\right)\)

900ml phai ko

900ml

a)

Gọi m KCl = a(gam)

Sau khi thêm : 

m KCl = a + 200.10% = a + 20(gam)
m dd = a + 200(gam)

Suy ra : 

C% = (a + 20) / (a +200) .100% = 15%

=> a = 11,76(gam)

b)

(Thiếu dữ kiện)

\(n_{OH^-}=0,01.0,1=0,001\left(mol\right)\)

\(pH=10\)

\(\Rightarrow\left[H^+\right]=10^{-10}\)

\(\Rightarrow\left[OH^-\right]=10^{-4}\)

\(\Leftrightarrow\dfrac{0,001}{0,1+V_{H_2O}}=10^{-4}\)

\(\Leftrightarrow V_{H_2O}=9,9\left(l\right)=9900\left(ml\right)\)

\(TC:\)

\(V_1+V_2=2\left(l\right)\)

\(m_{dd_{NaOH\left(3\%\right)}}=1.05V_1\left(g\right)\)

\(m_{NaOH\left(3\%\right)}=1.05V_1\cdot3\%=0.0315V_1\left(g\right)\)

\(m_{dd_{NaOH\left(10\%\right)}}=1.12V_2\left(g\right)\)

\(m_{NaOH\left(10\%\right)}=1.12V_2\cdot10\%=0.112V_2\left(g\right)\)

\(m_{NaOH\left(8\%\right)}=2000\cdot1.1\cdot8\%=176\left(g\right)\)

\(\Leftrightarrow0.0315V_1+0.112V_2=176\left(2\right)\)

\(\left(1\right),\left(2\right):V_1=596\left(ml\right),V_2=1404\left(ml\right)\)

Gọi V dd NaOH 3% = a(lít) ; V dd NaOH 10% = b(lít)

Ta có :  a + b = 2(1)

Áp dụng CT : m dd = D.V

m dd NaOH 3% = a.1,05.1000 = 1050a(gam)

m dd NaOH 10% = b.1,12.1000 = 1120b(gam)

m dd NaOH 8% = 2.1,1.1000 = 2200(gam)

Sau khi pha : 

m NaOH = 1050a.3% + 1120b.10% = 2200.8%(2)

Từ (1)(2) suy ra a = 0,596(lít) = 596(ml) ; b = 1,404(lít) = 1404(ml)

Ta có: n_NaOH = 0,2.0,25 = 0,05 (mol)

Gọi: V_H2O = a (l)

\(\Rightarrow\dfrac{0,05}{a+0,2}=0,1\) \(\Rightarrow a=0,3\left(l\right)=300\left(ml\right)\)

 \(V_{H_2O} = x(l) \\ 200ml=0,2l \\ n_{NaOH}=0,2 . 0,25=0,05(mol) \\ \Rightarrow C_{M\ NaOH\ 0,1M}=\frac{0,05}{a+0,2}=0,1(M) \\ Giải\ phương\ trình: x=0,3(l)=300(ml) \)