Tính C% của dung dịch thu được:

Ta có: m_{d d NaOH(1)}=V.D=500.1,2=600(g)

V_{d d NaOH(1)}=500ml=0,5 (lít)

=> n_NaOH(1)=C_M.V=2.0,5=1 (mol)

=> m_NaOH(1)=n_NaOH.M=1.40=40(gam)

Ta có: m_{d d NaOH(2)}=V.D=300.1,1=330(g)

V_{d d NaOH(2)}=300ml=0,3 (lít)

=> n_NaOH(2)=C_M.V=0,5.0,3=0,15(mol)

=> m_NaOH(2)=n.M=0,15.40=6(gam)

=> m_{NaOH mới}=m_NaOH(1) + m_NaOH(2)=40+6=46(gam)

m_{d d NaOH mới}=m_{d d NaOH(1)} + m_{d d NaOH(2)}=600+330=930(gam)

=> \(C\%_{ddsauphanung}=\dfrac{m_{NaOHmới}.100\%}{m_{ddNaOHmoi}}=\dfrac{46.100}{930}\approx4,95\left(\%\right)\)

Tính C_M của dung dịch thu được :

Ta có: V_{d d NaOH mới}=V_{d d NaOH(1)} + V_{d d NaOH(2)}=0,5+0,3=0,8(lít)

n_{d d NaOH mới}= n _{d d NaOH(1)} + n _{d d NaOH(2)}= 1 + 0,15=1,15(mol)

=> \(C_M=\dfrac{n}{V}=\dfrac{1,15}{0,8}\approx1,44\left(M\right)\)

a) n HCl = 0,2.2,5 = 0,5 mol

b) m HCl =200.7,3% = 14,6 gam

n HCl = 14,6/36,5 = 0,4 mol

c) m NaOH = 300.40% = 120 gam

n NaOH = 120/40 = 3(mol)

d) n NaOH = 0,5.0,5 = 0,25 mol

a) \(n_{Na}=\dfrac{11,5}{23}=0,5\left(mol\right)\)

\(n_{NaOH}=\dfrac{8\%.500}{40}=1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

0,5---------------->0,5------->0,25

\(\Sigma n_{NaOH}=0,5+1=1,5\left(mol\right)\)

\(m_{ddsaupu}=11,5+500-0,25.2=511\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{1,5.40}{511}.100=11,74\%\)

b) Gọi thể tích dung dịch X cần tìm là V

 \(n_{H^+}=V.1+V.0,5.1=2V\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

Ta có : \(n_{H^+}=n_{OH^-}=1,5\left(mol\right)\)

=> 2V=1,5

=> V=0,75(lít)

Dd X chỉ chứa 1 chất tan.

→ Pư vừa đủ, chất tan là BaCl₂.

Ta có: \(n_{Ba\left(OH\right)_2}=0,3.0,05=0,015\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

_____0,015_____0,03___0,015 (mol)

\(\Rightarrow a=C_{M_{HCl}}=\dfrac{0,03}{0,5}=0,06\left(M\right)\)

m _{chất rắn khan} = m_BaCl2 = 0,015.208 = 3,12 (g)

\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)

\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)

\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)

a, Ta có: \(n_{KOH}=\frac{0,56}{56}=0,01\left(mol\right)\)

\(KOH\rightarrow K^++OH^-\)

0,01 __________ 0,01 (mol)

\(\Rightarrow\left[OH^-\right]=\frac{0,01}{0,5}=0,02M\)

\(\Rightarrow\left[H^+\right]=5.10^{-13}M\)

\(\Rightarrow pH=-log\left[H^+\right]\approx12,3\)

b, Ta có: \(n_{H^+}=2n_{H_2SO_4}=0,08\left(mol\right)\)

\(n_{OH^-}=n_{NaOH}=0,059\left(mol\right)\)

PT ion: \(H^++OH^-\rightarrow H_2O\)

_____0,059 ← 0,059 (mol)

⇒ OH^- dư. \(n_{OH^-\left(dư\right)}=0,021\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]=\frac{0,021}{0,7}=0,03M\)

\(\Rightarrow\left[H^+\right]\approx3,33.10^{-13}\)

\(\Rightarrow pH\approx12,5\)

Bạn tham khảo nhé!

a, Ta có: \(n_{KOH}=0,005\left(mol\right)\)

\(\Rightarrow n_{OH^-}=n_{KOH}=0,005\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]=\frac{0,005}{0,5}=0,01M\)

\(\Rightarrow\left[H^+\right]=10^{-12}M\Rightarrow pH=12\)

b, Ta có: \(n_{H^+}=2n_{H_2SO_4}=0,06\left(mol\right)\)

\(n_{OH^-}=n_{NaOH}=0,05\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,05 ← 0,05 (mol)

⇒ H⁺ dư. \(n_{H^+\left(dư\right)}=0,01\left(mol\right)\Rightarrow\left[H^+\right]=0,0125M\)

\(\Rightarrow pH=-log\left[H^+\right]\approx1,9\)

Bạn tham khảo nhé!

a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)

b, \(n_{KOH\left(trong300mlA\right)}=0,3.0,3=0,09\left(mol\right)\)

Gọi: V_H2O = a (l)

\(\Rightarrow C_{M_A}=0,2=\dfrac{0,09}{a+0,3}\Rightarrow a=0,15\left(l\right)=150\left(ml\right)\)