Answer:
The correct answer is 0.277 M HCl
Explanation:
NaOH is a strong base and HCl is a strong acid. NaOH reacts with HCl to form a salt (NaCl) and water via a neutralization reaction, as follows:
NaOH + HCl → NaCl + H₂O
At the equivalence point, the moles of NaOH added is equal to the moles of HCl present in the solution to be titrated. Moreover, the moles can be calculated as the product of the concentration of base or acid (Cb or Ca) and the volume (Vb and Va), as follows:
moles NaOH = moles HCl
Cb x Vb = Ca x Va
Thus, we calculate the concentration of HCl (Ca), as follows:
Ca = (Cb x Vb)/Va = (0.146 M x 47.5 mL)/(25.0 mL)= 0.277 M
I am not sure about this answer 194
Answer:
Pb⁺²(aq) + CO₃⁻²(aq) → PbCO₃ (s)
In net ionic equation we cancel the ions that have equal moles on both sides so Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.
Explanation:
Net ionic equation:
In net ionic equation we only write the ions that are involved in reaction. If the system have same moles of ions in initial and final stages we cancel them as they have the same amount and are present in ionic form in the reaction medium. To formulate an ionic equation we just cancel the ions which have the same moles in initial and final stages.
Chemical equation:
Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + NaNO₃ (aq)
Balanced chemical equation:
In a balanced chemical equation we write the reactants and products in molecular form with number of moles.
Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + 2NaNO₃ (aq)
Ionic equation:
In ionic equation we write the equation in ionic form. It involves all the ions which will produce when we add any ionic compound in reaction medium.
Pb⁺² +2NO₃⁻¹ + CO₃⁻² + 2Na⁺¹ → PbCO3 (s) + 2NO₃⁻¹ (aq) + 2Na⁺¹ (aq)
Net ionic equation
In net ionic equation we cancel the ions that have equal moles on both sides. As we can see in the above ionic equation that Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.
Pb⁺²(aq) + CO₃⁻²(aq) → PbCO₃ (s)
A. physical change took place during the experiment. (plato users)
