answer:To solve this problem, we can use the concept of molarity and the balanced chemical equation for the neutralization reaction between HCl and NaOH: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. We can use the formula: moles of HCl = moles of NaOH We are given the volume and molarity of NaOH, so we can calculate the moles of NaOH: moles of NaOH = Molarity × Volume moles of NaOH = 0.2 M × 0.025 L (converting 25 mL to liters) moles of NaOH = 0.005 moles Now we can find the moles of HCl required to neutralize the NaOH: moles of HCl = moles of NaOH moles of HCl = 0.005 moles Next, we can find the volume of 0.1 M HCl required using the formula: Volume = moles / Molarity Volume = 0.005 moles / 0.1 M Volume = 0.05 L Converting liters to milliliters: Volume = 0.05 L × 1000 mL/L Volume = 50 mL So, 50 mL of 0.1 M HCl is required to neutralize 25 mL of 0.2 M NaOH.

answer:To find the molarity of the hydrochloric acid solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH): HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) From the balanced equation, we can see that the mole ratio between HCl and NaOH is 1:1. First, we need to find the moles of NaOH in the reaction: moles of NaOH = Molarity × Volume moles of NaOH = 0.15 M × 0.035 L (converting mL to L) moles of NaOH = 0.00525 moles Since the mole ratio between HCl and NaOH is 1:1, the moles of HCl that reacted are also 0.00525 moles. Now we can find the molarity of the HCl solution: Molarity of HCl = moles of HCl / Volume of HCl Molarity of HCl = 0.00525 moles / 0.025 L (converting mL to L) Molarity of HCl = 0.21 M So, the molarity of the hydrochloric acid solution is 0.21 M.

answer:To solve this problem, we can use the concept of moles and the balanced chemical equation for the neutralization reaction between NaOH and HCl: NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. First, let's find the moles of HCl in the 25 mL solution with a concentration of 0.2 M: moles of HCl = volume (L) × concentration (M) moles of HCl = 0.025 L × 0.2 M = 0.005 moles Since 1 mole of NaOH reacts with 1 mole of HCl, we need 0.005 moles of NaOH to neutralize the HCl completely. Now, we can find the volume of the 0.1 M NaOH solution needed: volume (L) = moles of NaOH / concentration (M) volume (L) = 0.005 moles / 0.1 M = 0.05 L Converting this to milliliters: volume (mL) = 0.05 L × 1000 mL/L = 50 mL So, 50 mL of the 0.1 M NaOH solution is needed to neutralize completely the 25 mL of the 0.2 M HCl solution.

answer:To calculate the volume of 0.2 M hydrochloric acid (HCl) required to completely neutralize 25 mL of 0.1 M sodium hydroxide (NaOH) solution, we can use the concept of moles and the balanced chemical equation for the reaction: NaOH + HCl → NaCl + H2O From the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. First, we need to find the moles of NaOH in the 25 mL solution: moles of NaOH = Molarity × Volume moles of NaOH = 0.1 M × 0.025 L (since 25 mL = 0.025 L) moles of NaOH = 0.0025 moles Since 1 mole of NaOH reacts with 1 mole of HCl, we need 0.0025 moles of HCl to neutralize the NaOH. Now, we can find the volume of 0.2 M HCl required: Volume of HCl = moles of HCl / Molarity of HCl Volume of HCl = 0.0025 moles / 0.2 M Volume of HCl = 0.0125 L To convert the volume to milliliters: Volume of HCl = 0.0125 L × 1000 mL/L Volume of HCl = 12.5 mL So, 12.5 mL of 0.2 M hydrochloric acid is required to completely neutralize 25 mL of 0.1 M sodium hydroxide solution.