When a 1.0 M solution of HCl is titrated with a 1.0 M solution of KOH, which is the approximate pH at the equivalent point?

1 answer:

4 0

The equivalence point is when the concentration of H⁺ in solution is equal to the concentration of OH⁻ in solution. Since H⁺ and OH⁻ react with each other to make water (H⁺(aq)+OH⁻(aq)→H₂O(l)) the pH at the equivalence point is 7 due to everything being neutralized. (The equivalence point only has a pH of 7 when a strong acid is being titrated with a strong base).

I hope this helps. Let me know if anything is unclear.

You might be interested in

Consider this reaction: Upper K Upper O Upper H + Upper H Upper B 4 right arrow Upper K Upper B r + Upper H Subscript 2 Baseline

Juliette [100K]

Hbr is the answer when I work it out that’s what I got.

4 0

2 years ago

Read 2 more answers

A 19.0 g piece of metal is heated to 99.0 °C and then placed in 150 mL of water at 21°C. The temperature of the water rose to 23

Anna35 [415]

Answer:

$0.8696\ \text{J/g}^{\circ}\text{C}$

Explanation:

$m_m$ = Mass of metal = 19 g

$c_m$ = Specific heat of the metal

$\Delta T_m$ = Temperature difference of the metal = $99-23=76^{\circ}\text{C}$

V = Volume of water = 150 mL = $150\ \text{cm}^3$

$\rho$ = Density of water = $1\ \text{g/cm}^3$

$c_w$ = Specific heat of the water = 4.186 J/g°C

$\Delta T_w$ = Temperature difference of the water = $23-21=2^{\circ}\text{C}$

Mass of water

$m_w= \rho V\\\Rightarrow m_w=1\times 150\\\Rightarrow m_w=150\ \text{g}$

Heat lost will be equal to the heat gained so we get

$m_mc_m\Delta T_m=m_wc_w\Delta T_w\\\Rightarrow c_m=\dfrac{m_wc_w\Delta T_w}{m_m\Delta T_m}\\\Rightarrow c_m=\dfrac{150\times 4.186\times 2}{19\times 76}\\\Rightarrow c_m=0.8696\ \text{J/g}^{\circ}\text{C}$